Question 1

(d^4 y)/(d x^4)+ sin (y^′ ′ ′)=0

Accepted Answer

Order 4; Degree not defined. By definition, the order of a differential equation is the order of the highest-order derivative occurring in it, and the degree is the power to which that highest-order derivative is raised — defined only when the equation is a polynomial equation in the derivatives y',y'',… The derivatives present are y''' and (d^4y)/(dx^4). The highest of these is the fourth derivative, so order =4. For the degree, test the polynomial requirement. Expanding the sine term, sin (y''')=y'''-((y''')^3)/(3!)+((y''')^5)/(5!)-, an infinite series in y'''; the equation therefore cannot be written as a polynomial equation in its derivatives, and no rearrangement removes the sine. Order =4; degree is not defined.

Question 2

y^′+5 y=0

Accepted Answer

Order 1; Degree 1. The order is the order of the highest derivative present; the degree is the power of that highest derivative, provided the equation is a polynomial equation in the derivatives. Here y'=dy/dx is the only derivative, so order =1. The equation y'+5y=0 is a polynomial in y' (and in y ), and y' occurs to the power 1. Order =1; degree =1.

Question 3

((d s)/(d t))^4+3 s (d^2 s)/(d t^2)=0

Accepted Answer

Order 2; Degree 1. The order is fixed by the highest-order derivative; the degree is the power of that derivative (not the largest exponent in the equation), provided the equation is polynomial in the derivatives. The derivatives present are ds/dt (order 1) and (d^2s)/(dt^2) (order 2). The highest-order derivative is (d^2s)/(dt^2), so order =2. The equation (ds/dt)^4+3s(d^2s)/(dt^2)=0 is a polynomial equation in ds/dt and (d^2s)/(dt^2), so the degree is defined. In it, (d^2s)/(dt^2) appears to the power 1; the exponent 4 belongs to the first derivative and is irrelevant to the degree. Order =2; degree =1.

Question 4

((d^2 y)/(d x^2))^2+ cos ((d y)/(d x))=0

Accepted Answer

Order 2; Degree not defined. The order is the order of the highest derivative; the degree is defined only if the equation is a polynomial equation in the derivatives. The derivatives occurring are dy/dx and (d^2y)/(dx^2), so order =2. The term cos (dy/dx) is a transcendental function of dy/dx: cos (dy/dx)=1-1/(2!)(dy/dx)^2+1/(4!)(dy/dx)^4-, which is not a polynomial in dy/dx. Hence the equation is not a polynomial equation in its derivatives. Order =2; degree is not defined.

Question 5

(d^2 y)/(d x^2)= cos 3 x+ sin 3 x

Accepted Answer

Order 2; Degree 1. First bring the equation to the form (expression in x,y and the derivatives) =0: (d^2y)/(dx^2)- cos 3x- sin 3x=0. The only derivative present is (d^2y)/(dx^2), so order =2. The trigonometric terms are functions of the independent variable x alone, not of any derivative, so they do not disturb the polynomial test: the left side is a polynomial in (d^2y)/(dx^2), which occurs to the power 1. Order =2; degree =1.

Question 6

(y^′ ′ ′)^2+(y^′ ′)^3+(y^′)^4+y^5=0

Accepted Answer

Order 3; Degree 2. The order comes from the highest-order derivative, and the degree is the power of that same derivative once the equation is polynomial in the derivatives. Derivatives present: y', y'', y'''. The highest is y''', so order =3. The equation (y''')^2+(y'')^3+(y')^4+y^5=0 is a polynomial equation in y',y'',y''', so the degree is defined, and it is the exponent of the highest-order derivative y''', namely 2. The larger exponents 3,4,5 attach to lower-order derivatives and to y itself, and are not the degree. Order =3; degree =2.

Question 7

y^′ ′ ′+2 y^′ ′+y^′=0

Accepted Answer

Order 3; Degree 1. The derivatives occurring are y', y'' and y'''; the highest-order one is y''', so order =3. The equation y'''+2y''+y'=0 is linear, hence certainly a polynomial equation in the derivatives, and y''' appears to the first power. Order =3; degree =1.

Question 8

y^′+y=e^x

Accepted Answer

Order 1; Degree 1. Write the equation as y'+y-e^x=0. The only derivative present is y', so order =1. The exponential e^x is a function of the independent variable x only — the polynomial test applies to the derivatives, and in them the left side is a polynomial with y' occurring to the power 1. Order =1; degree =1.

Question 9

y^′ ′+(y^′)^2+2 y=0

Accepted Answer

Order 2; Degree 1. The derivatives present are y' and y''; the highest-order derivative is y'', so order =2. The equation y''+(y')^2+2y=0 is a polynomial equation in y' and y'', so the degree is defined. It is the power of the highest-order derivative y'', which is 1; the square belongs to y' and does not make the degree 2. Order =2; degree =1.

Question 10

y^′ ′+2 y^′+ sin y=0

Accepted Answer

Order 2; Degree 1. The derivatives present are y' and y'', so the highest-order derivative is y'' and order =2. For the degree, the requirement is that the equation be a polynomial equation in the derivatives. The term sin y involves the dependent variable y itself, not any derivative, so it does not violate that requirement; in y' and y'' the equation y''+2y'+ sin y=0 is a polynomial, with y'' occurring to the power 1. Order =2; degree =1.

Question 11

The degree of the differential equation ((d^2 y)/(d x^2))^3+((d y)/(d x))^2+ sin ((d y)/(d x))+1=0 is (A) 3 (B) 2 (C) 1 (D) not defined

Accepted Answer

D. The degree of a differential equation is the power of the highest-order derivative, and it exists only when the equation is a polynomial equation in the derivatives. The highest-order derivative here is (d^2y)/(dx^2), which appears as ((d^2y)/(dx^2))^3; that alone would suggest degree 3. But the equation also contains sin (dy/dx), and sin (dy/dx)=dy/dx-1/(3!)(dy/dx)^3+1/(5!)(dy/dx)^5- is not a polynomial in dy/dx. So the equation is not a polynomial equation in its derivatives, and the degree fails to exist — the presence of a non-polynomial term in any derivative destroys the degree, whatever the exponent on the highest one. Hence the correct option is (D) not defined.

Question 12

The order of the differential equation 2 x^2 (d^2 y)/(d x^2)-3 (d y)/(d x)+y=0 is (A) 2 (B) 1 (C) 0 (D) not defined

Accepted Answer

A. The order of a differential equation is the order of the highest-order derivative occurring in it. (The coefficients — here 2x^2 and -3 — play no part in deciding the order.) In 2x^2(d^2y)/(dx^2)-3dy/dx+y=0 the derivatives present are dy/dx (order 1) and (d^2y)/(dx^2) (order 2); the highest of these is the second derivative. Hence the order is 2, and the correct option is (A) 2.

Question 13

y=e^x+1: y^′ ′-y^′=0

Accepted Answer

To verify a proposed solution, differentiate it as many times as the equation requires and substitute into the equation; the function is a solution if the equation reduces to an identity. Given y=e^x+1. Differentiating with respect to x (the derivative of the constant 1 is 0 ): y^′=e^x. Differentiating once more: y^′′=e^x. Substituting in the left-hand side of the given equation: y^′′-y^′=e^x-e^x=0, which is the right-hand side. Hence y=e^x+1 satisfies y^′′-y^′=0, so it is a solution of the given differential equation.

Question 14

y=x^2+2 x+ C: y^′-2 x-2=0

Accepted Answer

Differentiate the given function and substitute into the equation. Given y=x^2+2x+ C, where C is an arbitrary constant. Differentiating with respect to x, and using d/dx( C )=0: y^′=2x+2. Note that the constant C disappears on differentiation, which is why the same derivative serves for every member of the family. Substituting in the left-hand side: y^′-2x-2=(2x+2)-2x-2=0, which is the right-hand side. Hence y=x^2+2x+ C is a solution of y^′-2x-2=0 for every value of C.

Question 15

y= cos x+ C: y^′+ sin x=0

Accepted Answer

Differentiate the given function and substitute into the equation. Given y= cos x+ C. Using d/dx( cos x)=- sin x and d/dx( C )=0: y^′=- sin x. The sign here is the step to watch: the derivative of cos x carries a minus sign. Substituting in the left-hand side: y^′+ sin x=- sin x+ sin x=0, which is the right-hand side. Hence y= cos x+ C is a solution of y^′+ sin x=0 for every value of C.

Question 16

y=√ 1+x^2: y^′=(x y)/(1+x^2)

Accepted Answer

Differentiate by the chain rule, then re-express the answer in terms of y itself. Given y=√ 1+x^2 =(1+x^2)^1/2. By the chain rule, y^′=1/2 (1+x^2)^-1/2 /dx(1+x^2)=1/2 (1+x^2)^-1/2· 2x= x √ 1+x^2. The right-hand side of the equation is written in terms of y, so multiply numerator and denominator by √ 1+x^2: y^′= x √ 1+x^2 · √ 1+x^2 √ 1+x^2 = x√ 1+x^2 1+x^2 =(x y)/(1+x^2), since y=√ 1+x^2. This is exactly the given differential equation, so y=√ 1+x^2 is a solution of y^′=xy/(1+x^2).

Question 17

y= A x: x y^′=y(x ≠ 0)

Accepted Answer

Differentiate the given function and substitute into the equation. Given y= A x, where A is an arbitrary constant. Differentiating with respect to x: y^′= A. Substituting in the left-hand side of the equation, for x≠ 0: x y^′=x· A = A x=y, which is the right-hand side. Hence y= A x is a solution of xy^′=y (x≠ 0) for every value of A; geometrically this is the family of straight lines through the origin.

Question 18

y=x sin x: x y^′=y+x √ x^2-y^2 (x ≠ 0 and x y or x<-y)

Accepted Answer

Differentiate by the product rule and then simplify the surd on the right-hand side using the Pythagorean identity. Given y=x sin x. By the product rule, y^′= sin x+x cos x. Left-hand side of the equation: x y^′=x sin x+x^2 cos x=y+x^2 cos x, since y=x sin x. Right-hand side: with y=x sin x, x^2-y^2=x^2-x^2 sin^2x=x^2(1- sin^2x)=x^2 cos^2x, using sin^2x+ cos^2x=1. Hence √ x^2-y^2 =√ x^2 cos^2x = x cos x. This is the step most easily got wrong: a square root is non-negative, so it is x cos x, not automatically x cos x. Taking the root as x cos x is legitimate precisely where x cos x≥ 0, and there y+x√ x^2-y^2 =y+x(x cos x)=y+x^2 cos x. The stated conditions x≠ 0 and x y or x<-y guarantee x^2 y^2, so the surd is real. Both sides equal y+x^2 cos x, so y=x sin x is a solution of xy^′=y+x√ x^2-y^2 on the intervals where x cos x≥ 0.

Question 19

x y= log y+C: y^′=(y^2)/(1-x y)(x y ≠ 1)

Accepted Answer

The function is given implicitly, so differentiate the relation as it stands, treating y as a function of x (implicit differentiation), and then solve the result for y^′. Given xy= log y+ C. Differentiating both sides with respect to x, using the product rule on the left and the chain rule on log y: y+x y^′=1/y y^′. Multiply throughout by y (which is legitimate since log y exists only for y 0, so y≠ 0 ): y^2+xy y^′=y^′. Collect the y^′ terms on one side: y^2=y^′-xy y^′=y^′(1-xy). Since xy≠ 1, we may divide by 1-xy: y^′=(y^2)/1-xy. This is precisely the given differential equation, so xy= log y+ C is an (implicit) solution of y^′=(y^2)/1-xy, xy≠ 1.

Question 20

y- cos y=x:(y sin y+ cos y+x) y^′=y

Accepted Answer

Differentiate the implicit relation, find y^′, and then use the relation itself to eliminate x from the left-hand side. Given y- cos y=x. Differentiating both sides with respect to x, treating y as a function of x: y^′-(- sin y)y^′=1, i.e. y^′(1+ sin y)=1, so y^′=1/(1+ sin y) (1+ sin y≠ 0). Now substitute in the left-hand side of the given equation: (y sin y+ cos y+x)y^′=(y sin y+ cos y+x)/(1+ sin y). The step that makes it work is replacing x by y- cos y from the given relation: y sin y+ cos y+x=y sin y+ cos y+(y- cos y)=y sin y+y=y(1+ sin y). Therefore (y sin y+ cos y+x)y^′=(y(1+ sin y))/(1+ sin y)=y, which is the right-hand side. Hence y- cos y=x is a solution of (y sin y+ cos y+x)y^′=y.

Question 21

x+y= tan^-1 y: y^2 y^′+y^2+1=0

Accepted Answer

Differentiate the implicit relation with respect to x and clear the fraction. Given x+y= tan^-1y. Differentiating both sides with respect to x, using d/dx( tan^-1y)=1/(1+y^2) y^′: 1+y^′=(y^′)/(1+y^2). Multiply both sides by 1+y^2 (never zero): (1+y^2)(1+y^′)=y^′, that is, 1+y^2+y^′+y^2y^′=y^′. Cancelling y^′ from both sides: y^2y^′+y^2+1=0. This is exactly the given differential equation, so x+y= tan^-1y is a solution of y^2y^′+y^2+1=0.

Question 22

y=√ a^2-x^2 x ∈(-a, a): x+y (d y)/(d x)=0(y ≠ 0)

Accepted Answer

Differentiate by the chain rule and express the derivative in terms of y. Given y=√ a^2-x^2 =(a^2-x^2)^1/2 for x∈(-a,a), so that a^2-x^2 0 and the root is real and positive. By the chain rule, dy/dx=1/2(a^2-x^2)^-1/2 /dx(a^2-x^2)=1/2(a^2-x^2)^-1/2(-2x)= -x √ a^2-x^2. The minus sign coming from d/dx(-x^2) is the step to be careful with. Since y=√ a^2-x^2 and y≠ 0 on (-a,a), dy/dx=-x/y. Substituting in the left-hand side of the equation: x+ydy/dx=x+y(-x/y)=x-x=0, which is the right-hand side. Hence y=√ a^2-x^2, x∈(-a,a), is a solution of x+ydy/dx=0, y≠ 0.

Question 23

The number of arbitrary constants in the general solution of a differential equation of fourth order are: (A) 0 (B) 2 (C) 3 (D) 4

Accepted Answer

D. Use the definition: the general solution of a differential equation is the solution containing as many arbitrary constants as the order of the equation. The order of a differential equation is the order of the highest derivative occurring in it. Recovering a function from its n th derivative requires n successive integrations, and each integration introduces one arbitrary constant; hence a differential equation of order n has a general solution with exactly n arbitrary constants. Here the order is 4, so the general solution contains 4 arbitrary constants. The correct option is ( D ) 4.

Question 24

The number of arbitrary constants in the particular solution of a differential equation of third order are: (A) 3 (B) 2 (C) 1 (D) 0

Accepted Answer

D. Distinguish the general solution from a particular solution. The general solution of a differential equation of order n contains n arbitrary constants. A particular solution is obtained from the general solution by assigning particular values to those arbitrary constants (for instance, from given initial or boundary conditions); once the values are fixed, no arbitrary constant remains. So for a third order equation the general solution has 3 arbitrary constants, but a particular solution has none, whatever the order. The correct option is ( D ) 0.

Question 25

(d y)/(d x)=(1- cos x)/(1+ cos x)

Accepted Answer

The variables are already separated, so this is a direct integration once the right-hand side is put in an integrable form. Use the half-angle identities 1- cos x = 2 sin^2x/2 and 1+ cos x = 2 cos^2x/2: dy/dx=(2 sin^2x/2)/(2 cos^2x/2)= tan^2x/2 The step most often missed: tan^2θ has no direct antiderivative, so convert it by tan^2θ= sec^2θ-1: dy=( sec^2x/2-1)dx Integrating both sides, and remembering that ∫ sec^2x/2 dx = 2 /2 because of the inner factor 12, y = 2 /2-x+C General solution: y = 2 /2 - x + C, valid wherever cos x ≠ -1.

Question 26

(d y)/(d x)=√ 4-y^2 (-2<y<2)

Accepted Answer

y=2 sin (x+ C ). Variables separable. For -2<y<2 we have 4-y^2 0, so we may divide by √ 4-y^2: dy √ 4-y^2 =dx Use the standard integral ∫ dy √ a^2-y^2 = sin^-1y/a+C with a=2; the restriction -2<y<2 is exactly what makes y/2 lie in (-1,1), the domain of sin^-1: sin^-1y/2=x+C Solving for y, y/2= sin (x+C) General solution: y = 2 sin (x+C), with C arbitrary.

Question 27

(d y)/(d x)+y=1(y ≠ 1)

Accepted Answer

Write the equation as dy/dx=1-y. Since y≠ 1 we may divide by 1-y and separate: dy/1-y=dx Integrating, and noting ∫ dy/1-y = - log 1-y (the minus sign from the inner derivative is the usual slip), - log 1-y = x + C 1 log 1-y = -x - C 1 1-y = e^ -C 1 e^-x ⇒ 1-y = A e^-x, A = ± e^ -C 1 ≠ 0 Hence y = 1 - Ae^-x. Writing C=-A, y = 1 + Ce^-x General solution: y = 1 + Ce^-x.

Question 28

sec^2 x tan y d x+ sec^2 y tan x d y=0

Accepted Answer

tan x tan y= C. The equation is variables separable once we divide throughout by tan x tan y (permissible where tan x ≠ 0, tan y ≠ 0 ): ( sec^2x)/( tan x) dx + ( sec^2y)/( tan y) dy = 0 Each integral is of the form ∫ (f'(t))/(f(t))dt = log f(t), since d/dx( tan x)= sec^2x: log tan x + log tan y = log C log tan x tan y = log C General solution: tan x tan y = C.

Question 29

(e^x+e^-x) d y-(e^x-e^-x) d x=0

Accepted Answer

Rearranging, dy/dx=(e^x-e^-x)/(e^x+e^-x) Observe that the numerator is exactly the derivative of the denominator: d/dx(e^x+e^-x)=e^x-e^-x. So this is again ∫ (f'(x))/(f(x))dx: y=∫ (e^x-e^-x)/(e^x+e^-x) dx = log (e^x+e^-x)+C No modulus sign is needed because e^x+e^-x 0 for every real x. General solution: y = log (e^x+e^-x)+C.

Question 30

(d y)/(d x)=(1+x^2)(1+y^2)

Accepted Answer

tan^-1 y=x+(x^3)/3+ C. The right-hand side already factors as a function of x times a function of y, so separate: dy/(1+y^2)=(1+x^2)dx Integrate, using ∫ dy/(1+y^2)= tan^-1y: tan^-1y = x + (x^3)/3+C General solution: tan^-1y = x + (x^3)/3+C, i.e. y = tan (x+(x^3)/3+C).

Question 31

y log y d x-x d y=0

Accepted Answer

Separate the variables (this needs x≠ 0, y 0 and log y ≠ 0, i.e. y≠ 1 ): y log y dx = x dy ⇒ dx/x=dy/(y log y) For the right side substitute t= log y, so dt = dy/y: ∫ dy/(y log y)=∫ dt/t= log t = log log y Therefore log log y = log x + log C log y = Cx General solution: y = e^Cx.

Question 32

x^5 (d y)/(d x)=-y^5

Accepted Answer

x^-4+y^-4= C. Separate the variables (valid for x≠ 0, y≠ 0 ): dy/(y^5)=-dx/(x^5) ∫ y^-5 dy = -∫ x^-5 dx Watch the two negative signs here, in the power rule and in the equation itself: (y^-4)/-4= -(x^-4)/-4+C 1 ⇒ -1/(4y^4)=1/(4x^4)+C 1 Multiplying through by -4 and writing C=-4C 1: 1/(x^4)+1/(y^4)=C General solution: x^-4+y^-4=C.

Question 33

(d y)/(d x)= sin^-1 x

Accepted Answer

y=x sin^-1 x+√ 1-x^2 + C. Direct integration: y=∫ sin^-1x dx. Integrate by parts with u= sin^-1x, dv = dx (the standard trick of taking the second function to be 1 ): ∫ sin^-1x· 1 dx = x sin^-1x - ∫ x √ 1-x^2 dx For the remaining integral put t = 1-x^2, dt = -2x dx: ∫ x √ 1-x^2 dx = -1/2∫ dt/(√ t ) = -√ 1-x^2 Hence y = x sin^-1x + √ 1-x^2 +C General solution: y = x sin^-1x + √ 1-x^2 +C, for -1≤ x≤ 1.

Question 34

e^x tan y d x+(1-e^x) sec^2 y d y=0

Accepted Answer

tan y= C (1-e^x). Divide throughout by (1-e^x) tan y to separate (needs e^x≠ 1, tan y ≠ 0 ): (e^x)/(1-e^x) dx + ( sec^2y)/( tan y) dy = 0 In the first integral the numerator is -1 times the derivative of the denominator, since d/dx(1-e^x)=-e^x; in the second, d/dy( tan y)= sec^2y: - log 1-e^x + log tan y = log C log ( tan y)/(1-e^x) = log C General solution: tan y = C(1-e^x).

Question 35

(x^3+x^2+x+1) (d y)/(d x)=2 x^2+x; y=1 when x=0

Accepted Answer

y=1/4 log [(x+1)^2(x^2+1)^3]-1/2 tan^-1 x+1. First factorise the coefficient by grouping: x^3+x^2+x+1 = x^2(x+1)+(x+1)=(x+1)(x^2+1) So dy = (2x^2+x)/((x+1)(x^2+1)) dx Resolve into partial fractions, using a linear numerator over the irreducible quadratic: (2x^2+x)/((x+1)(x^2+1))=A/(x+1)+(Bx+C)/(x^2+1) 2x^2+x = A(x^2+1)+(Bx+C)(x+1) Put x=-1: 2-1 = 2A ⇒ A= 12. Comparing coefficients of x^2: 2 = A+B ⇒ B = 32. Comparing constants: 0 = A + C ⇒ C = - 12. y=∫[1/2·1/(x+1)+3/2 /(x^2+1)-1/2·1/(x^2+1)]dx y = 1/2 log x+1 + 3/4 log (x^2+1)-1/2 tan^-1x + K (the 34 comes from 32· 12, since ∫ (x dx)/(x^2+1)= 12 log (x^2+1) ). Apply y=1 when x=0: 1 = 12 log 1 + 34 log 1 - 12 tan^-10 + K = K, so K=1. Particular solution: y = 1/4[2 log x+1 + 3 log (x^2+1) - 2 tan^-1x] + 1.

Question 36

x(x^2-1) (d y)/(d x)=1; y=0 when x=2

Accepted Answer

y=1/2 log ((x^2-1)/(x^2))-1/2 log 3/4. Separate and factorise the denominator: dy = dx/(x(x^2-1)) = dx/(x(x-1)(x+1)) Partial fractions with three distinct linear factors: 1/(x(x-1)(x+1))=A/x+B/x-1+C/(x+1) 1 = A(x-1)(x+1)+Bx(x+1)+Cx(x-1) Put x=0: 1 = -A ⇒ A=-1. Put x=1: 1 = 2B ⇒ B= 12. Put x=-1: 1 = 2C ⇒ C= 12. y = - log x + 1/2 log x-1 + 1/2 log x+1 + K = 1/2 log (x^2-1)/(x^2) + K Apply y=0 when x=2: 0 = 1/2 log 3/4+K ⇒ K = 1/2 log 4/3 Particular solution: y = 1/2 log (4(x^2-1))/(3x^2).

Question 37

cos ((d y)/(d x))=a(a ∈ R ); y=1 when x=0

Accepted Answer

cos (y-2/x)=a. The equation is not separable as printed; first solve it for dy/dx. Since cos (dy/dx)=a, a real derivative exists only if -1≤ a≤ 1; taking the principal value of the inverse cosine, dy/dx= cos^-1a The right-hand side is a constant, so integrating, y = x cos^-1a + C Apply y=1 when x=0: 1 = 0 + C, so C=1. y = x cos^-1a + 1 Rearranging for the form usually quoted, for x≠ 0: y-1/x= cos^-1a ⇒ cos (y-1/x)=a Particular solution: y = 1 + x cos^-1a, equivalently cos (y-1/x)=a, valid for a ≤ 1.

Question 38

(d y)/(d x)=y tan x; y=1 when x=0

Accepted Answer

y= sec x. Variables separable (assume y≠ 0 ): dy/y= tan x dx Integrate, using ∫ tan x dx = log sec x: log y = log sec x + log C y = C sec x Apply y=1 when x=0: 1 = C sec 0 = C, so C=1. Particular solution: y = sec x, on the interval containing x=0 where cos x 0, i.e. -(π)/2<x<(π)/2.

Question 39

Find the equation of a curve passing through the point (0,0) and whose differential equation is y^′=e^x sin x.

Accepted Answer

2 y-1=e^x( sin x- cos x). Here dy/dx=e^x sin x, so y=∫ e^x sin x dx. Evaluate by parts twice and solve for the integral. Let I=∫ e^x sin x dx; taking sin x as the first function, I = e^x sin x - ∫ e^x cos x dx = e^x sin x - [e^x cos x + ∫ e^x sin x dx] I = e^x sin x - e^x cos x - I ⇒ 2I = e^x( sin x- cos x) y = (e^x( sin x- cos x))/2+C The curve passes through (0,0): 0 = (e^0(0-1))/2+C = -1/2+C ⇒ C=1/2 Equation of the curve: y = (e^x( sin x - cos x)+1)/2, i.e. 2y-1 = e^x( sin x - cos x).

Question 40

For the differential equation x y (d y)/(d x)=(x+2)(y+2), find the solution curve passing through the point (1, -1).

Accepted Answer

y-x+2= log (x^2(y+2)^2). The equation is variables separable after dividing by x(y+2) (needs x≠ 0, y≠ -2 ): y/(y+2) dy = (x+2)/x dx Neither side can be integrated directly; split each fraction first (this is the step to get right): (1-2/(y+2))dy = (1+2/x)dx Integrating, y - 2 log y+2 = x + 2 log x + C y - x = 2 log x + 2 log y+2 + C = 2 log x(y+2) + C The curve passes through (1,-1), where x=1 and y+2=1: -1-1 = 2 log 1 + C ⇒ C = -2 Solution curve: y - x + 2 = 2 log x(y+2), equivalently y-x+2 = log [x^2(y+2)^2].

Question 41

Find the equation of a curve passing through the point (0, -2) given that at any point (x, y) on the curve, the product of the slope of its tangent and y coordinate of the point is equal to the x coordinate of the point.

Accepted Answer

y^2-x^2=4. The slope of the tangent at (x,y) is dy/dx. Translating the condition "slope × y -coordinate = x -coordinate": y dy/dx=x Separate and integrate: y dy = x dx ⇒ (y^2)/2=(x^2)/2+C 1 y^2-x^2=C, C = 2C 1 The curve passes through (0,-2): (-2)^2-0^2=C ⇒ C=4 Equation of the curve: y^2-x^2=4, a rectangular hyperbola (the branch through (0,-2), i.e. y=-√ x^2+4 ).

Question 42

At any point (x, y) of a curve, the slope of the tangent is twice the slope of the line segment joining the point of contact to the point (-4, -3). Find the equation of the curve given that it passes through (-2, 1).

Accepted Answer

(x+4)^2=y+3. The slope of the segment joining (x,y) to (-4,-3) is (y-(-3))/(x-(-4))=(y+3)/(x+4). The condition "slope of tangent is twice this" gives the differential equation dy/dx=(2(y+3))/(x+4) Separate (valid for y≠ -3, x≠ -4 ) and integrate: dy/(y+3)=(2 dx)/(x+4) log y+3 = 2 log x+4 + log A = log [A(x+4)^2] y+3 = A(x+4)^2 The curve passes through (-2,1): 1+3 = A(-2+4)^2=4A ⇒ A=1 Equation of the curve: y+3 = (x+4)^2, a parabola.

Question 43

The volume of spherical balloon being inflated changes at a constant rate. If initially its radius is 3 units and after 3 seconds it is 6 units. Find the radius of balloon after t seconds.

Accepted Answer

(63 t+27)^1/3. Let r be the radius at time t and V=4/3π r^3 the volume. "Volume changes at a constant rate" means dV/dt=k ⇒ d/dt(4/3π r^3)=k Integrating with respect to t: 4/3π r^3=kt + C Use r=3 at t=0: 4/3π(27)=36π = C. Use r=6 at t=3: 4/3π(216)=288π = 3k+36π, so 3k = 252π and k = 84π. Hence 4/3π r^3=84π t + 36π ⇒ r^3=63t+27 Radius after t seconds: r = (63t+27)^1/3 units (check: t=0 gives 3, t=3 gives 216^1/3=6 ).

Question 44

In a bank, principal increases continuously at the rate of r % per year. Find the value of r if Rs 100 double itself in 10 years ( log e 2=0.6931).

Accepted Answer

6.93\%. Let P be the principal at time t years. "Increases continuously at the rate of r % per year" means dP/dt=r/100P Separate and integrate: dP/P=r/100 dt ⇒ log P = rt/100+C 1 ⇒ P = P 0e^rt/100 Here P 0=100 (the value at t=0 ). The money doubles in 10 years, so P=200 at t=10: 200 = 100 e^10r/100 ⇒ e^r/10=2 ⇒ r/10= log e2 = 0.6931 r = 6.931 Hence r = 6.931 % per year.

Question 45

In a bank, principal increases continuously at the rate of 5\% per year. An amount of Rs 1000 is deposited with this bank, how much will it worth after 10 years (e^0.5=1.648).

Accepted Answer

Rs 1648. With P the principal at time t years, continuous growth at 5 % per year gives dP/dt=5/100P = P/20 Separating and integrating, dP/P=dt/20 ⇒ log P = t/20+C 1 ⇒ P = P 0e^t/20 Given P 0=1000 at t=0, so P = 1000 e^t/20. At t=10: P = 1000 e^10/20=1000 e^0.5=1000(1.648) After 10 years the deposit is worth Rs 1648.

Question 46

In a culture, the bacteria count is 1,00,000. The number is increased by 10\% in 2 hours. In how many hours will the count reach 2,00,000, if the rate of growth of bacteria is proportional to the number present?

Accepted Answer

(2 log 2)/( log (11/10)). Let y be the bacteria count at time t hours. "Rate of growth proportional to the number present" gives dy/dt=ky Separating and integrating, dy/y=k dt ⇒ log y = kt + C 1 ⇒ y = y 0e^kt with y 0=1, 00, 000. A 10 % increase in 2 hours means y = 1, 10, 000 at t=2: 1, 10, 000 = 1, 00, 000 e^2k ⇒ e^2k=11/10 ⇒ k = 1/2 log 11/10 Now require y = 2, 00, 000, i.e. e^kt=2: kt = log 2 ⇒ t = ( log 2)/k=(2 log 2)/( log 11/10) Numerically, t = (2(0.6931))/0.0953≈ 14.5. The count reaches 2, 00, 000 in t = (2 log 2)/( log (11/10))≈ 14.5 hours.

Question 47

The general solution of the differential equation (d y)/(d x)=e^x+y is (A) e^x+e^-y= C (B) e^x+e^y= C (C) e^-x+e^y= C (D) e^-x+e^-y= C

Accepted Answer

A. Use the law of exponents to factor the right-hand side into a function of x times a function of y, which makes the equation separable: dy/dx=e^x+y=e^xe^y e^-y dy = e^x dx Integrating both sides, and noting ∫ e^-ydy = -e^-y (the sign here decides between the options): -e^-y=e^x+C 1 e^x+e^-y=-C 1= C Hence the correct option is (A) e^x+e^-y= C.

Question 48

(x^2+x y) d y=(x^2+y^2) d x

Accepted Answer

Write the equation as dy/dx=(x^2+y^2)/(x^2+xy). Numerator and denominator are both homogeneous of degree 2, so the right side is a function of y/x alone and the equation is homogeneous. Substitute y=vx, so that dy/dx=v+xdv/dx: v+xdv/dx=(x^2+v^2x^2)/(x^2+vx^2)=(1+v^2)/(1+v). Therefore xdv/dx=(1+v^2)/(1+v)-v=(1+v^2-v-v^2)/(1+v)=1-v/(1+v). Separating the variables, (1+v)/1-v dv=dx/x. The left integrand is improper in v; rewrite 1+v=-(1-v)+2, so that (1+v)/1-v=-1+2/1-v. Integrating, -v-2 log 1-v = log x + C. Put v=y/x and use log 1- yx = log x-y - log x: - yx-2 log x-y +2 log x = log x + C, log ((x-y)^2)/( x )=- yx- C. Exponentiating, the general solution is (x-y)^2e^y/x= C 1 x, equivalently (x-y)^2= C 1 x e^-y/x, with C 1 an arbitrary constant and x≠0.

Question 49

y^′=(x+y)/x

Accepted Answer

y=x log x + C x. Here dy/dx=(x+y)/x=1+ yx, a function of y/x alone, so the equation is homogeneous. Substitute y=vx, so dy/dx=v+xdv/dx: v+xdv/dx=1+v ⇒ xdv/dx=1. Separating the variables, dv=dx/x, and integrating, v= log x + C. Replacing v= yx, the general solution is y=x( log x + C), x≠0.

Question 50

(x-y) d y-(x+y) d x=0

Accepted Answer

Rearranging, (x-y) dy=(x+y) dx, i.e. dy/dx=(x+y)/x-y. Numerator and denominator are homogeneous of degree 1, so the equation is homogeneous. Put y=vx, dy/dx=v+xdv/dx: v+xdv/dx=(x+vx)/x-vx=(1+v)/1-v, xdv/dx=(1+v-v(1-v))/1-v=(1+v^2)/1-v. Separating the variables and splitting the left side into the two standard integrals, 1-v/(1+v^2) dv=dx/x ⇒ /(1+v^2)-∫(v dv)/(1+v^2)= /x. The second integral needs t=1+v^2, giving 12 log (1+v^2). Hence tan^-1v- 12 log (1+v^2)= log x + C. With v= yx, 12 log (1+(y^2)/(x^2))= 12 log (x^2+y^2)- log x, and the log x terms cancel: tan^-1 yx- 12 log (x^2+y^2)= C. So the general solution is tan^-1( yx)= 12 log (x^2+y^2)+ C.

Question 51

(x^2-y^2) d x+2 x y d y=0

Accepted Answer

x^2+y^2= C x. Write the equation as dy/dx=(y^2-x^2)/2xy. Every term is homogeneous of degree 2, so the equation is homogeneous. Put y=vx, dy/dx=v+xdv/dx: v+xdv/dx=(v^2x^2-x^2)/(2vx^2)=(v^2-1)/2v, xdv/dx=(v^2-1-2v^2)/2v=-(1+v^2)/2v. Separating the variables, 2v/(1+v^2) dv=-dx/x. The left side is exactly d/dv log (1+v^2), so integrating gives log (1+v^2)=- log x + C. Substituting v= yx, log (x^2+y^2)/(x^2)+ log x = C ⇒ log (x^2+y^2)/( x )= C. Exponentiating, the general solution is x^2+y^2= C 1x, a family of circles through the origin with centres on the x -axis ( x≠0 ).

Question 52

x^2 (d y)/(d x)=x^2-2 y^2+x y

Accepted Answer

Divide by x^2 (x≠0): dy/dx=1-2( yx)^2+ yx, a function of y/x alone, so the equation is homogeneous. Put y=vx, dy/dx=v+xdv/dx: v+xdv/dx=1-2v^2+v ⇒ xdv/dx=1-2v^2. Separating the variables, dv/(1-2v^2)=dx/x. For the left side, factor out the 2 so as to reach the standard form /(a^2-t^2)=1/2a log (a+t)/a-t: /(1-2v^2)= 12 /((1/(√2))^2-v^2)= 12·(√2)/2 log (1/(√2)+v)/(1/(√2)-v) =1/(2√2) log 1+√2 v 1-√2 v. Hence 1/(2√2) log (1+√2 v)/(1-√2 v) = log x + C. Putting v= yx and multiplying numerator and denominator inside the logarithm by x, the general solution is 1/(2√2) log (x+√2 y)/(x-√2 y) = log x + C.

Question 53

x d y-y d x=√ x^2+y^2 d x

Accepted Answer

y+√ x^2+y^2 = C x^2. Write the equation as dy/dx= y+√ x^2+y^2 x; both parts of the numerator are homogeneous of degree 1, so the equation is homogeneous. Put y=vx, dy/dx=v+xdv/dx. The step to watch is the surd: √ x^2+v^2x^2 = x √ 1+v^2. Take x 0 (for x<0 only the sign of the arbitrary constant changes), so v+xdv/dx= vx+x√ 1+v^2 x =v+√ 1+v^2, xdv/dx=√ 1+v^2. Separating the variables, dv √ 1+v^2 =dx/x. Using ∫ dv √ v^2+a^2 = log v+√ v^2+a^2, log v+√ 1+v^2 = log x + C. Exponentiating, v+√ 1+v^2 = C 1x. Substituting v= yx and √ 1+v^2 = √ x^2+y^2 x, y/x+ √ x^2+y^2 x = C 1x. The general solution is y+√ x^2+y^2 = C 1x^2.

Question 54

x cos (y/x)+y sin (y/x) y d x= y sin (y/x)-x cos (y/x) x d y

Accepted Answer

Write the equation as dy/dx=(y x cos ( yx)+y sin ( yx) )/(x y sin ( yx)-x cos ( yx) ). Numerator and denominator are homogeneous of degree 2, so the equation is homogeneous. Put y=vx, dy/dx=v+xdv/dx. The numerator becomes vx(x cos v+vx sin v)=vx^2( cos v+v sin v) and the denominator x(vx sin v-x cos v)=x^2(v sin v- cos v). Hence v+xdv/dx=(v( cos v+v sin v))/(v sin v- cos v), xdv/dx=(v cos v+v^2 sin v-v(v sin v- cos v))/(v sin v- cos v)=(2v cos v)/(v sin v- cos v). Note how the v^2 sin v terms cancel, leaving only 2v cos v. Separating the variables, (v sin v- cos v)/(v cos v) dv=(2 dx)/x ⇒ ( tan v- 1v)dv=(2 dx)/x. Integrating, with ∫ tan v dv=- log cos v, - log cos v - log v =2 log x + C, log 1/(v cos v) = log x^2+ C ⇒ v x^2 cos v= C 1. Substituting v= yx, so that v x^2=xy, the general solution is xy cos ( yx)= C.

Question 55

x (d y)/(d x)-y+x sin (y/x)=0

Accepted Answer

x[1- cos (y/x)]= C sin (y/x). Write the equation as dy/dx=(y-x sin ( yx))/x= yx- sin ( yx), a function of y/x alone, so the equation is homogeneous. Put y=vx, dy/dx=v+xdv/dx: v+xdv/dx=v- sin v ⇒ xdv/dx=- sin v. Separating the variables, cosec v dv=-dx/x. Using ∫ cosec v dv= log tan v2, log tan v2 =- log x + C ⇒ log x tan v2 = C. Exponentiating and substituting v= yx, the general solution is x tan (y/2x)= C 1, x≠0.

Question 56

y d x+x log (y/x) d y-2 x d y=0

Accepted Answer

c y= log y/x -1. Collect the dy terms: y dx= 2x-x log ( yx) dy, so dy/dx=y/(x 2- log ( yx) ), which depends only on y/x; the equation is homogeneous. Put y=vx, dy/dx=v+xdv/dx: v+xdv/dx=vx/(x(2- log v))=v/(2- log v), xdv/dx=(v-v(2- log v))/(2- log v)=(v( log v-1))/(2- log v). Separating the variables, (2- log v)/(v( log v-1)) dv=dx/x. Substitute t= log v-1, so dt=dv/v and 2- log v=1-t. The left side becomes ∫1-t/t dt=∫( 1t-1)dt= log t -t. Hence log log v-1 -( log v-1)= log x + C, log log v-1 = log x + log v+ C 1. With v= yx the two logarithms on the right combine: log x + log yx= log y. So log log yx-1 = log y + C 1. Exponentiating, the general solution is log ( yx)-1= C y, i.e. log ( yx)=1+ C y.

Question 57

(1+e^x/y) d x+e^x/y(1-x/y) d y=0

Accepted Answer

y e^x/y+x= C. Every occurrence of the variables is through x/y, so this equation is homogeneous but in the x/y form; the right substitution is x=vy, not y=vx. Solving for dx/dy, dx/dy=-(e^x/y(1- xy))/(1+e^x/y). Put x=vy, so dx/dy=v+ydv/dy: v+ydv/dy=-(e^v(1-v))/(1+e^v), ydv/dy=(-e^v+ve^v-v(1+e^v))/(1+e^v)=(-e^v+ve^v-v-ve^v)/(1+e^v)=-(v+e^v)/(1+e^v). Separating the variables, ((1+e^v)dv)/(v+e^v)=-dy/y. The numerator on the left is exactly the derivative of the denominator, so integrating gives log v+e^v =- log y + C ⇒ y(v+e^v)= C 1. Substituting v= xy, y( xy+e^x/y)= C 1. The general solution is x+y e^x/y= C, y≠0.

Question 58

(x+y) d y+(x-y) d x=0; y=1 when x=1

Accepted Answer

log (x^2+y^2)+2 tan^-1 y/x=(π)/2+ log 2. Write the equation as dy/dx=y-x/(x+y); numerator and denominator are homogeneous of degree 1, so the equation is homogeneous. Put y=vx, dy/dx=v+xdv/dx: v+xdv/dx=vx-x/(x+vx)=v-1/(1+v), xdv/dx=(v-1-v(1+v))/(1+v)=-(1+v^2)/(1+v). Separating the variables and splitting the left side, (1+v)/(1+v^2) dv=-dx/x ⇒ /(1+v^2)+∫(v dv)/(1+v^2)=- /x, tan^-1v+ 12 log (1+v^2)=- log x + C. With v= yx, 12 log (1+(y^2)/(x^2))= 12 log (x^2+y^2)- log x, and the log x terms cancel: tan^-1 yx+ 12 log (x^2+y^2)= C. Apply the condition y=1 when x=1: tan^-11+ 12 log 2= C ⇒ C=(π)/4+ 12 log 2. The particular solution is tan^-1( yx)+ 12 log (x^2+y^2)=(π)/4+ 12 log 2, or equivalently log (x^2+y^2)+2 tan^-1( yx)=(π)/2+ log 2.

Question 59

x^2 d y+(x y+y^2) d x=0; y=1 when x=1

Accepted Answer

y+2 x=3 x^2 y. Write the equation as dy/dx=-(xy+y^2)/(x^2); every term is homogeneous of degree 2, so the equation is homogeneous. Put y=vx, dy/dx=v+xdv/dx: v+xdv/dx=-(vx^2+v^2x^2)/(x^2)=-(v+v^2), xdv/dx=-2v-v^2=-v(v+2). Separating the variables, dv/(v(v+2))=-dx/x. Resolve the left side into partial fractions, 1/(v(v+2))= 12( 1v-1/(v+2)). Integrating, 12 log v/(v+2) =- log x + C ⇒ log v/(v+2) =-2 log x + C 1. Exponentiating, v/(v+2)=( A)/(x^2). With v= yx, (y/x)/( yx+2)=y/(y+2x)=( A)/(x^2) ⇒ x^2y= A(y+2x). Apply y=1 when x=1: 1= A(1+2), so A= 13. Then 3x^2y=y+2x, i.e. y(3x^2-1)=2x. The particular solution is y=2x/(3x^2-1), x≠±1/(√3).

Question 60

[x sin^2(y/x)-y] d x+x d y=0; y=(π)/4 when x=1

Accepted Answer

cot (y/x)= log e x. Solve for the derivative: x dy=[y-x sin^2( yx)]dx, so dy/dx= yx- sin^2( yx), a function of y/x alone; the equation is homogeneous. Put y=vx, dy/dx=v+xdv/dx: v+xdv/dx=v- sin^2v ⇒ xdv/dx=- sin^2v. Separating the variables, cosec^2v dv=-dx/x. Since ∫ cosec^2v dv=- cot v, integrating gives - cot v=- log x + C ⇒ cot v= log x + C 1. Apply y=(π)/4 when x=1, so v=(π)/4 and cot (π)/4=1: 1= log 1+ C 1= C 1. The particular solution is cot ( yx)= log x +1= log ex.

Question 61

(d y)/(d x)-y/x+ cosec (y/x)=0; y=0 when x=1

Accepted Answer

cos (y/x)= log e x. The equation is dy/dx= yx- cosec ( yx), a function of y/x alone, so it is homogeneous. Put y=vx, dy/dx=v+xdv/dx: v+xdv/dx=v- cosec v ⇒ xdv/dx=- cosec v. Separating the variables, sin v dv=-dx/x. Integrating, - cos v=- log x + C ⇒ cos v= log x + C 1. Apply y=0 when x=1, so v= 01=0 and cos 0=1: 1= log 1+ C 1= C 1. The particular solution is cos ( yx)= log x +1= log ex.

Question 62

2 x y+y^2-2 x^2 (d y)/(d x)=0; y=2 when x=1

Accepted Answer

y=(2 x)/(1- log x )(x ≠ 0, x ≠ e). Solve for the derivative: 2x^2dy/dx=2xy+y^2, so dy/dx=(2xy+y^2)/(2x^2), homogeneous of degree 2 in numerator and denominator. Put y=vx, dy/dx=v+xdv/dx: v+xdv/dx=(2vx^2+v^2x^2)/(2x^2)=v+(v^2)/2, xdv/dx=(v^2)/2. Separating the variables, (2 dv)/(v^2)=dx/x ⇒ - 2v= log x + C. Substituting v= yx, -2x/y= log x + C. Apply y=2 when x=1: -2/2= log 1+ C, so C=-1. Then -2x/y= log x -1 ⇒ 2x/y=1- log x. The particular solution is y=2x/(1- log x ), x≠0, x≠ e.

Question 63

A homogeneous differential equation of the from (d x)/(d y)=h(x/y) can be solved by making the substitution. (A) y=v x (B) v=y x (C) x=v y (D) x=v

Accepted Answer

C. In dx/dy=h( xy) the dependent variable is x and the independent variable is y, and the right side is a function of the ratio xy. The substitution must turn that ratio into a single new variable, so we set the ratio equal to v: xy=v ⇒ x=vy, dx/dy=v+ydv/dy. The equation then becomes v+ydv/dy=h(v), i.e. dv/(h(v)-v)=dy/y, which is variables-separable. (The substitution y=vx would leave the equation still containing both x and v unseparated, so it is the wrong one here.) Hence the correct option is (C) x=vy.

Question 64

Which of the following is a homogeneous differential equation? (A) (4 x+6 y+5) d y-(3 y+2 x+4) d x=0 (B) (x y) d x-(x^3+y^3) d y=0 (C) (x^3+2 y^2) d x+2 x y d y=0 (D) y^2 d x+(x^2-x y-y^2) d y=0

Accepted Answer

D. A first order differential equation M(x,y) dx+ N(x,y) dy=0 is homogeneous when M and N are homogeneous functions of the same degree, i.e. F(λ x,λ y)=λ^n F(x,y). Test each option by degree. (A) (4x+6y+5) dy-(3y+2x+4) dx=0: the constants 5 and 4 are of degree 0 while 4x+6y and 3y+2x are of degree 1, so neither coefficient is homogeneous. Not homogeneous. (B) (xy) dx-(x^3+y^3)dy=0: xy is of degree 2, x^3+y^3 of degree 3. Degrees differ. Not homogeneous. (C) (x^3+2y^2)dx+2xy dy=0: within the first coefficient x^3 is of degree 3 and 2y^2 of degree 2, so that coefficient is not homogeneous at all. Not homogeneous. (D) y^2 dx+(x^2-xy-y^2)dy=0: M=y^2 and N=x^2-xy-y^2 are both homogeneous of degree 2, since replacing (x,y) by (λ x,λ y) multiplies each by λ^2. Equivalently dy/dx=(-y^2)/(x^2-xy-y^2) is a function of yx alone. Hence the correct option is (D) y^2 dx+(x^2-xy-y^2)dy=0.

EXERCISE 9.1 1–12 (part 1 of 10)

Exercise 1

Exercise 2

Exercise 3

Exercise 4

Exercise 5

Exercise 6

Exercise 7

Exercise 8

Exercise 9

Exercise 10

Exercise 11

Exercise 12