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NCERT Exemplar · Class 12 Mathematics Differential Equations

77 questions · 77 still being checked

EXERCISE 9.3 61–70 (part 7 of 8)

  1. Choose the correct answer from the given four options in each of the Exercises from $\displaystyle 34$ to $\displaystyle 75$ (M.C.Q)

    Exercise 61

    The general solution of dydx=2xex2−y\displaystyle \frac{d y}{d x}=2 x e^{x^2-y} is :
    (A)
    ex2−y=c\displaystyle e^{x^2-y}=c
    (B)
    e−y+ex2=c\displaystyle e^{-y}+e^{x^2}=c
    (C)
    ey=ex2+c\displaystyle e^y=e^{x^2}+c
    (D)
    ex2+y=c\displaystyle e^{x^2+y}=c

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    NCERT’s answer
    C
    (C) \(\displaystyle e^y = e^{x^2} + c\). Split the exponent and separate the variables.\[\frac{dy}{dx} = 2x\,e^{x^2}\,e^{-y} \] \[e^{y}\,dy = 2x\,e^{x^2}\,dx \] \[\int e^{y}\,dy = \int e^{x^2}\,d\left(x^2\right) \] \[e^{y} = e^{x^2} + c \]

  2. Exercise 62

    The curve for which the slope of the tangent at any point is equal to the ratio of the abcissa to the ordinate of the point is :
    (A)
    an ellipse
    (B)
    parabola
    (C)
    circle
    (D)
    rectangular hyperbola

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    NCERT’s answer
    D
    (D) rectangular hyperbola. Slope equals abscissa over ordinate.\[\frac{dy}{dx} = \frac{x}{y} \] \[y\,dy = x\,dx \] \[\frac{y^2}{2} = \frac{x^2}{2} + c \] \[x^2 - y^2 = -2c \]

  3. Exercise 63

    The general solution of the differential equation dydx=ex22+xy\displaystyle \frac{d y}{d x}=e^{\frac{x^2}{2}}+x y is :
    (A)
    y=ce−x22\displaystyle y=c e^{\frac{-x^2}{2}}
    (B)
    y=cex22\displaystyle y=c e^{\frac{x^2}{2}}
    (C)
    y=(x+c)ex22\displaystyle y=(x+c) e^{\frac{x^2}{2}}
    (D)
    y=(c−x)ex22\displaystyle y=(c-x) e^{\frac{x^2}{2}}

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    NCERT’s answer
    C
    (C) \(\displaystyle y=(x+c)\,e^{x^2/2}\). Linear equation; use the integrating factor.\[\frac{dy}{dx} - xy = e^{x^2/2} \] \[\text{I.F.} = e^{-\int x\,dx} = e^{-x^2/2} \] \[\frac{d}{dx}\left(y\,e^{-x^2/2}\right) = e^{x^2/2}\cdot e^{-x^2/2} = 1 \] \[y\,e^{-x^2/2} = x + c \] \[y = (x + c)\,e^{x^2/2} \]

  4. Exercise 64

    The solution of the equation (2y−1)dx−(2x+3)dy=0\displaystyle (2 y-1) d x-(2 x+3) d y=0 is :
    (A)
    2x−12y+3=k\displaystyle \frac{2 x-1}{2 y+3}=k
    (B)
    2y+12x−3=k\displaystyle \frac{2 y+1}{2 x-3}=k
    (C)
    2x+32y−1=k\displaystyle \frac{2 x+3}{2 y-1}=k
    (D)
    2x−12y−1=k\displaystyle \frac{2 x-1}{2 y-1}=k

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    NCERT’s answer
    C
    (C) \(\displaystyle \dfrac{2x+3}{2y-1}=k\). Separate the variables and integrate.\[(2y-1)\,dx = (2x+3)\,dy \] \[\frac{dx}{2x+3} = \frac{dy}{2y-1} \] \[\frac{1}{2}\log|2x+3| = \frac{1}{2}\log|2y-1| + \frac{1}{2}\log|k| \] \[\frac{2x+3}{2y-1} = k \]

  5. Exercise 65

    The differential equation for which y=acos⁡x+bsin⁡x\displaystyle y=a \cos x+b \sin x is a solution, is :
    (A)
    d2ydx2+y=0\displaystyle \frac{d^2 y}{d x^2}+y=0
    (B)
    d2ydx2−y=0\displaystyle \frac{d^2 y}{d x^2}-y=0
    (C)
    d2ydx2+(a+b)y=0\displaystyle \frac{d^2 y}{d x^2}+(a+b) y=0
    (D)
    d2ydx2+(a−b)y=0\displaystyle \frac{d^2 y}{d x^2}+(a-b) y=0

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    NCERT’s answer
    A
    (A) \(\displaystyle \dfrac{d^2y}{dx^2}+y=0\)Differentiate twice.\[\frac{dy}{dx} = -a\sin x + b\cos x \] \[\frac{d^2y}{dx^2} = -a\cos x - b\sin x = -y \] \[\frac{d^2y}{dx^2} + y = 0 \]

  6. Exercise 66

    The solution of dydx+y=e−x,y(0)=0\displaystyle \frac{d y}{d x}+y=e^{-x}, y(0)=0 is :
    (A)
    y=e−x(x−1)\displaystyle y=e^{-x}(x-1)
    (B)
    y=xex\displaystyle y=x e^x
    (C)
    y=xe−x+1\displaystyle y=x e^{-x}+1
    (D)
    y=xe−x\displaystyle y=x e^{-x}

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    NCERT’s answer
    D
    (D) \(\displaystyle y=xe^{-x}\)Linear equation; integrating factor:\[\text{I.F.} = e^{\int 1\,dx} = e^{x} \] \[\frac{d}{dx}\left(y\,e^{x}\right) = e^{-x}\cdot e^{x} = 1 \] \[y\,e^{x} = x + c \] \[y(0)=0 \;\Rightarrow\; 0 = 0 + c \;\Rightarrow\; c = 0 \] \[y = xe^{-x} \]

  7. Exercise 67

    The order and degree of the differential equation
    (d3ydx3)2−3d2ydx2+2(dydx)4=y4\left(\frac{d^3 y}{d x^3}\right)^2-3 \frac{d^2 y}{d x^2}+2\left(\frac{d y}{d x}\right)^4=y^4 are :
    (A)
    1\displaystyle 1, 4\displaystyle 4
    (B)
    3\displaystyle 3, 4\displaystyle 4
    (C)
    2\displaystyle 2, 4\displaystyle 4
    (D)
    3\displaystyle 3, 2\displaystyle 2

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    NCERT’s answer
    D
    (D) order \(\displaystyle 3\), degree \(\displaystyle 2\)The highest derivative present, and its power (the equation is a polynomial in the derivatives):\[\text{highest derivative} = \frac{d^3y}{dx^3} \;\Rightarrow\; \text{order} = 3 \] \[\left(\frac{d^3y}{dx^3}\right)^{2} \;\Rightarrow\; \text{degree} = 2 \]

  8. Exercise 68

    The order and degree of the differential equation [1+(dydx)2]=d2ydx2\displaystyle \left[1+\left(\frac{d y}{d x}\right)^2\right]=\frac{d^2 y}{d x^2} are :
    (A)
    2,32\displaystyle 2, \frac{3}{2}
    (B)
    2\displaystyle 2, 3\displaystyle 3
    (C)
    2\displaystyle 2, 1\displaystyle 1
    (D)
    3\displaystyle 3, 4\displaystyle 4

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    NCERT’s answer
    C
    (C) order \(\displaystyle 2\), degree \(\displaystyle 1\)The equation is a polynomial in the derivatives.\[\text{highest derivative} = \frac{d^2y}{dx^2} \;\Rightarrow\; \text{order} = 2 \] \[\left(\frac{d^2y}{dx^2}\right)^{1} \;\Rightarrow\; \text{degree} = 1 \]

  9. Exercise 69

    The differential equation of the family of curves y2=4a(x+a)\displaystyle y^2=4 a(x+a) is :
    (A)
    y2=4dydx(x+dydx)\displaystyle y^2=4 \frac{d y}{d x}\left(x+\frac{d y}{d x}\right)
    (B)
    2ydydx=4a\displaystyle 2 y \frac{d y}{d x}=4 a
    (C)
    yd2ydx2+(dydx)2=0\displaystyle y \frac{d^2 y}{d x^2}+\left(\frac{d y}{d x}\right)^2=0
    (D)
    2xdydx+y(dydx)2−y\displaystyle 2 x \frac{d y}{d x}+y\left(\frac{d y}{d x}\right)^2-y

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    (D) \(\displaystyle 2x\dfrac{dy}{dx}+y\left(\dfrac{dy}{dx}\right)^2-y=0\)
    Eliminate \(\displaystyle a\) (option (D) is read as an expression equal to zero).
    \[y^2 = 4a(x+a) \]
    \[2y\,y' = 4a \;\Rightarrow\; a = \frac{y\,y'}{2} \]
    \[y^2 = 2y\,y'\left(x + \frac{y\,y'}{2}\right) = 2xy\,y' + y^2 (y')^2 \]
    \[y = 2x\,y' + y\,(y')^2 \]
    \[2x\frac{dy}{dx} + y\left(\frac{dy}{dx}\right)^2 - y = 0 \]
    (C)
    also holds, since \(\displaystyle yy' = 2a\) is constant:
    \[(yy')' = 0 \;\Rightarrow\; y\,y'' + (y')^2 = 0 \]
    But it is second order; one constant \(\displaystyle a\) needs a first-order equation.

  10. Exercise 70

    Which of the following is the general solution of d2ydx2−2dydx+y=0\displaystyle \frac{d^2 y}{d x^2}-2 \frac{d y}{d x}+y=0 ?
    (A)
    y=(Ax+B)ex\displaystyle y=(\mathrm{A} x+\mathrm{B}) \mathrm{e}^x
    (B)
    y=(Ax+B)e−x\displaystyle y=(\mathrm{A} x+\mathrm{B}) e^{-x}
    (C)
    y=Aex+Be−x\displaystyle y=\mathrm{A} e^x+\mathrm{B} e^{-x}
    (D)
    y=Acos⁡x+Bsin⁡x\displaystyle y=\mathrm{A} \cos x+\mathrm{B} \sin x

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    NCERT’s answer
    A
    (A) \(\displaystyle y=(Ax+B)e^{x}\)Substitute; two constants in a second-order equation make it the general solution.\[y = (Ax+B)\,e^{x} \] \[y' = (Ax+A+B)\,e^{x} \] \[y'' = (Ax+2A+B)\,e^{x} \] \[y'' - 2y' + y = e^{x}\left[(Ax+2A+B) - 2(Ax+A+B) + (Ax+B)\right] = 0 \]