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NCERT Exemplar · Class 12 Mathematics Differential Equations

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EXERCISE 9.3 1–10 (part 1 of 8)

  1. Exercise 1

    Find the solution of dydx=2y−x\displaystyle \frac{d y}{d x}=2^{y-x}.

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    NCERT’s answer
    \(\displaystyle 2^{-x}-2^{-y}=k\)
    Separate the variables. \[\frac{dy}{dx} = \frac{2^{y}}{2^{x}} \] \[2^{-y}\,dy = 2^{-x}\,dx \] \[\int 2^{-y}\,dy = \int 2^{-x}\,dx \] \[-\frac{2^{-y}}{\ln 2} = -\frac{2^{-x}}{\ln 2} + c \] \[2^{-x} - 2^{-y} = C \qquad (C = c\ln 2) \] Answer: \(\displaystyle 2^{-x} - 2^{-y} = C\)

  2. Exercise 2

    Find the differential equation of all non vertical lines in a plane.

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    NCERT’s answer
    \(\displaystyle \frac{d^2 y}{d x^2}=0\)
    Every non-vertical line has a slope and an intercept. \[y = mx + c \] Two arbitrary constants \(\displaystyle m, c\), so differentiate twice. \[\frac{dy}{dx} = m \] \[\frac{d^2y}{dx^2} = 0 \] Answer: \(\displaystyle \dfrac{d^2y}{dx^2} = 0\)

  3. Exercise 3

    Given that dydx=e−2y\displaystyle \frac{d y}{d x}=e^{-2 y} and y=0\displaystyle y=0 when x=5\displaystyle x=5. Find the value of x\displaystyle x when y=3\displaystyle y=3.

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    NCERT’s answer
    \(\displaystyle \frac{e^6+9}{2}\)
    \[\frac{dy}{dx} = e^{-2y} \] \[e^{2y}\,dy = dx \] \[\frac{e^{2y}}{2} = x + c \] Use \(\displaystyle y = 0\) when \(\displaystyle x = 5\). \[\frac12 = 5 + c \;\Rightarrow\; c = -\frac92 \] \[e^{2y} = 2x - 9 \] Put \(\displaystyle y = 3\). \[e^{6} = 2x - 9 \] \[x = \frac{e^{6} + 9}{2} \] Answer: \(\displaystyle x = \dfrac{e^{6}+9}{2}\)

  4. Exercise 4

    Solve the differential equation (x2−1)dydx+2xy=1x2−1\displaystyle \left(x^2-1\right) \frac{d y}{d x}+2 x y=\frac{1}{x^2-1}.

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    NCERT’s answer
    \(\displaystyle y\left(x^2-1\right)=\frac{1}{2} \log \left(\left|\frac{x-1}{x+1}\right|\right)+k\)
    Divide by \(\displaystyle x^2 - 1\) to get the linear form. \[\frac{dy}{dx} + \frac{2x}{x^2-1}\,y = \frac{1}{(x^2-1)^2} \] \[\text{I.F.} = e^{\int \frac{2x}{x^2-1}dx} = e^{\ln|x^2-1|} \] The sign of \(\displaystyle x^2-1\) multiplies both sides alike, so take \(\displaystyle \text{I.F.} = x^2 - 1\). \[y\,(x^2-1) = \int \frac{1}{(x^2-1)^2}\,(x^2-1)\,dx = \int \frac{dx}{x^2-1} \] \[\frac{1}{x^2-1} = \frac12\left(\frac{1}{x-1} - \frac{1}{x+1}\right) \] \[y\,(x^2-1) = \frac12 \ln\left|\frac{x-1}{x+1}\right| + C \] Answer: \(\displaystyle y\,(x^2-1) = \dfrac12 \ln\left|\dfrac{x-1}{x+1}\right| + C\)

  5. Exercise 5

    Solve the differential equation dydx+2xy=y\displaystyle \frac{d y}{d x}+2 x y=y

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    NCERT’s answer
    \(\displaystyle y=c . e^{x-x^2}\)
    \[\frac{dy}{dx} = y\,(1 - 2x) \] \[\frac{dy}{y} = (1 - 2x)\,dx \] \[\ln|y| = x - x^2 + c \] \[y = A\,e^{\,x - x^2} \qquad (A = \pm e^{c}) \] Answer: \(\displaystyle y = A\,e^{x - x^2}\), \(\displaystyle A\) an arbitrary constant

  6. Exercise 6

    Find the general solution of dydx+ay=emx\displaystyle \frac{d y}{d x}+a y=e^{m x}

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    NCERT’s answer
    \(\displaystyle (a+m) y=e^{m x}+c e^{-a x}\)
    Linear equation in \(\displaystyle y\). \[\text{I.F.} = e^{\int a\,dx} = e^{ax} \] \[y\,e^{ax} = \int e^{(m+a)x}\,dx \] For \(\displaystyle m + a \neq 0\): \[y\,e^{ax} = \frac{e^{(m+a)x}}{m+a} + C \] \[y = \frac{e^{mx}}{m+a} + C\,e^{-ax} \] For \(\displaystyle m + a = 0\) the integrand is \(\displaystyle 1\): \[y\,e^{ax} = x + C \] \[y = (x + C)\,e^{-ax} \] Answer: \(\displaystyle y = \dfrac{e^{mx}}{m+a} + C\,e^{-ax}\) for \(\displaystyle m \neq -a\); \(\displaystyle y = (x + C)\,e^{-ax}\) for \(\displaystyle m = -a\)

  7. Exercise 7

    Solve the differential equation dydx+1=ex+y\displaystyle \frac{d y}{d x}+1=e^{x+y}

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    NCERT’s answer
    \(\displaystyle (x-c) e^{x+y}+1=0\)
    Put \(\displaystyle v = x + y\). \[\frac{dv}{dx} = 1 + \frac{dy}{dx} \] The equation becomes \[\frac{dv}{dx} = e^{v} \] \[e^{-v}\,dv = dx \] \[-e^{-v} = x + c \] \[x + e^{-(x+y)} = C \qquad (C = -c) \] Answer: \(\displaystyle x + e^{-(x+y)} = C\)

  8. Exercise 8

    Solve: ydx−xdy=x2ydx\displaystyle y d x-x d y=x^2 y d x.

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    NCERT’s answer
    \(\displaystyle y=k x e^{\frac{-x^2}{2}}\)
    \[y\,dx - x\,dy = x^2 y\,dx \] \[x\,dy = y\,(1 - x^2)\,dx \] \[\frac{dy}{y} = \left(\frac1x - x\right)dx \] \[\ln|y| = \ln|x| - \frac{x^2}{2} + c \] \[y = A\,x\,e^{-x^2/2} \qquad (A = \pm e^{c}) \] Answer: \(\displaystyle y = A\,x\,e^{-x^2/2}\), \(\displaystyle A\) an arbitrary constant

  9. Exercise 9

    Solve the differential equation dydx=1+x+y2+xy2\displaystyle \frac{d y}{d x}=1+x+y^2+x y^2, when y=0,x=0\displaystyle y=0, x=0.

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    NCERT’s answer
    \(\displaystyle y=\tan \left(x+\frac{x^2}{2}\right)\)
    Factorise the right side, then separate the variables. \[\frac{dy}{dx} = (1+x)(1+y^2) \] \[\frac{dy}{1+y^2} = (1+x)\,dx \] \[\tan^{-1} y = x + \frac{x^2}{2} + C \] \[x=0,\ y=0 \;\Rightarrow\; C = 0 \] \[y = \tan\!\left(x + \frac{x^2}{2}\right) \] Answer: \(\displaystyle y = \tan\left(x + \dfrac{x^2}{2}\right)\)

  10. Exercise 10

    Find the general solution of (x+2y3)dydx=y\displaystyle \left(x+2 y^3\right) \frac{d y}{d x}=y.

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    NCERT’s answer
    \(\displaystyle x=y\left(y^2+c\right)\)
    Take \(\displaystyle x\) as the dependent variable; the equation is linear in \(\displaystyle x\). \[\frac{dx}{dy} = \frac{x + 2y^3}{y} \] \[\frac{dx}{dy} - \frac{1}{y}\,x = 2y^2 \] \[\text{I.F.} = e^{-\int \frac{dy}{y}} = \frac{1}{y} \] \[\frac{x}{y} = \int 2y\,dy = y^2 + C \] \[x = y^3 + Cy \] Answer: \(\displaystyle x = y^3 + Cy\)