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NCERT Exemplar · Class 12 Mathematics Differential Equations

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EXERCISE 9.3 11–20 (part 2 of 8)

  1. Exercise 11

    If y(x)\displaystyle y(x) is a solution of (2+sin⁡x1+y)dydx=−cos⁡x\displaystyle \left(\frac{2+\sin x}{1+y}\right) \frac{d y}{d x}=-\cos x and y(0)=1\displaystyle \mathrm{y}(0)=1, then find the value of y(π2)\displaystyle y\left(\frac{\pi}{2}\right).

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    NCERT’s answer
    \(\displaystyle \frac{1}{3}\)
    Separate the variables. \[\frac{dy}{1+y} = -\frac{\cos x}{2+\sin x}\,dx \] \[\ln|1+y| = -\ln(2+\sin x) + \ln C \] \[(1+y)(2+\sin x) = C \] \[x=0,\ y=1:\quad C = (1+1)(2+0) = 4 \] \[x = \frac{\pi}{2}:\quad (1+y)(2+1) = 4 \] \[y\!\left(\frac{\pi}{2}\right) = \frac{4}{3} - 1 = \frac{1}{3} \] Answer: \(\displaystyle y\left(\dfrac{\pi}{2}\right) = \dfrac{1}{3}\)

  2. Exercise 12

    If y(t)\displaystyle y(t) is a solution of (1+t)dydt−ty=1\displaystyle (1+t) \frac{d y}{d t}-t y=1 and y(0)=−1\displaystyle y(0)=-1, then show that y(1)=−12.y(1)=-\frac{1}{2} .

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    Write in linear form. \[\frac{dy}{dt} - \frac{t}{1+t}\,y = \frac{1}{1+t} \] \[-\frac{t}{1+t} = -1 + \frac{1}{1+t} \] \[\int -\frac{t}{1+t}\,dt = -t + \ln(1+t) \] \[\text{I.F.} = e^{-t + \ln(1+t)} = e^{-t}(1+t) \] \[y\,e^{-t}(1+t) = \int e^{-t}(1+t)\cdot\frac{1}{1+t}\,dt = -e^{-t} + C \] \[t=0,\ y=-1:\quad -1 = -1 + C \;\Rightarrow\; C = 0 \] \[y(1+t) = -1 \;\Rightarrow\; y = -\frac{1}{1+t} \] \[y(1) = -\frac{1}{1+1} = -\frac{1}{2} \] Answer: \(\displaystyle y(1) = -\dfrac{1}{2}\)

  3. Exercise 13

    Form the differential equation having y=(sin⁡−1x)2+Acos−1x+B\displaystyle y=\left(\sin ^{-1} x\right)^2+\mathrm{Acos}^{-1} x+\mathrm{B}, where A and B are arbitrary constants, as its general solution.

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    NCERT’s answer
    \(\displaystyle \left(1-x^2\right) \frac{d^2 y}{d x^2}-x \frac{d y}{d x}-2=0\)
    Two arbitrary constants, so differentiate twice. \[\frac{dy}{dx} = \frac{2\sin^{-1}x}{\sqrt{1-x^2}} - \frac{A}{\sqrt{1-x^2}} \] \[\sqrt{1-x^2}\,\frac{dy}{dx} = 2\sin^{-1}x - A \] \[\sqrt{1-x^2}\,\frac{d^2y}{dx^2} - \frac{x}{\sqrt{1-x^2}}\,\frac{dy}{dx} = \frac{2}{\sqrt{1-x^2}} \] \[(1-x^2)\,\frac{d^2y}{dx^2} - x\,\frac{dy}{dx} = 2 \] Answer: \(\displaystyle (1-x^2)\dfrac{d^2y}{dx^2} - x\dfrac{dy}{dx} = 2\)

  4. Exercise 14

    Form the differential equation of all circles which pass through origin and whose centres lie on y\displaystyle y-axis.

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    NCERT’s answer
    \(\displaystyle \left(x^2-y^2\right) \frac{d y}{d x}-2 x y=0\)
    Centre \(\displaystyle (0,k)\), and the origin lies on the circle, so the radius is \(\displaystyle |k|\). \[x^2 + (y-k)^2 = k^2 \] \[x^2 + y^2 = 2ky \] \[2x + 2y\,\frac{dy}{dx} = 2k\,\frac{dy}{dx} \] \[k = \frac{x + y\,\dfrac{dy}{dx}}{\dfrac{dy}{dx}} \] \[x^2 + y^2 = 2y\,\frac{x + y\,\dfrac{dy}{dx}}{\dfrac{dy}{dx}} \] \[(x^2 + y^2)\frac{dy}{dx} = 2xy + 2y^2\,\frac{dy}{dx} \] \[(x^2 - y^2)\,\frac{dy}{dx} = 2xy \] Answer: \(\displaystyle (x^2 - y^2)\dfrac{dy}{dx} - 2xy = 0\)

  5. Exercise 15

    Find the equation of a curve passing through origin and satisfying the differential equation (1+x2)dydx+2xy=4x2\displaystyle \left(1+x^2\right) \frac{d y}{d x}+2 x y=4 x^2.

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    NCERT’s answer
    \(\displaystyle y=\frac{4 x^3}{3\left(1+x^2\right)}\)
    The left side is the derivative of \(\displaystyle (1+x^2)\,y\). \[\frac{d}{dx}\Big[(1+x^2)\,y\Big] = 4x^2 \] \[(1+x^2)\,y = \frac{4x^3}{3} + C \] \[x=0,\ y=0 \;\Rightarrow\; C = 0 \] \[y = \frac{4x^3}{3(1+x^2)} \] Answer: \(\displaystyle y = \dfrac{4x^3}{3(1+x^2)}\)

  6. Exercise 16

    Solve : x2dydx=x2+xy+y2\displaystyle x^2 \frac{d y}{d x}=x^2+x y+y^2.

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    NCERT’s answer
    \(\displaystyle \tan ^{-1}\left(\frac{y}{x}\right)=\log |x|+c\)
    Homogeneous equation; put \(\displaystyle y = vx\). \[\frac{dy}{dx} = 1 + \frac{y}{x} + \frac{y^2}{x^2} \] \[\frac{dy}{dx} = v + x\,\frac{dv}{dx} \] \[v + x\,\frac{dv}{dx} = 1 + v + v^2 \] \[\frac{dv}{1+v^2} = \frac{dx}{x} \] \[\tan^{-1} v = \ln|x| + C \] \[\tan^{-1}\frac{y}{x} = \ln|x| + C \] Answer: \(\displaystyle \tan^{-1}\dfrac{y}{x} = \ln|x| + C\)

  7. Exercise 17

    Find the general solution of the differential equation (1+y2)+(x−etan⁡−1y)dydx=0\displaystyle \left(1+y^2\right)+\left(x-e^{\tan -1 y}\right) \frac{d y}{d x}=0.

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    NCERT’s answer
    \(\displaystyle 2 x e^{\tan ^{-1} y}=e^{2 \tan ^{-1} y}+c\)
    Linear in \(\displaystyle x\) with \(\displaystyle y\) as the variable: \[\frac{dx}{dy}+\frac{x}{1+y^2}=\frac{e^{\tan^{-1}y}}{1+y^2} \] \[\text{I.F.}=e^{\int\frac{dy}{1+y^2}}=e^{\tan^{-1}y} \] \[x\,e^{\tan^{-1}y}=\int\frac{e^{2\tan^{-1}y}}{1+y^2}\,dy \] Put \(\displaystyle t=\tan^{-1}y\), so \(\displaystyle dt=\frac{dy}{1+y^2}\): \[x\,e^{t}=\int e^{2t}\,dt=\tfrac12e^{2t}+C \] \[x=\tfrac12e^{\tan^{-1}y}+C\,e^{-\tan^{-1}y} \] Answer: \(\displaystyle x\,e^{\tan^{-1}y}=\tfrac12e^{2\tan^{-1}y}+C\)

  8. Exercise 18

    Find the general solution of y2dx+(x2−xy+y2)dy=0\displaystyle y^2 d x+\left(x^2-x y+y^2\right) d y=0.

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    NCERT’s answer
    \(\displaystyle \tan ^{-1}\left(\frac{x}{y}\right)+\log y=c\)
    Homogeneous in \(\displaystyle x\) and \(\displaystyle y\); put \(\displaystyle x=vy\): \[\frac{dx}{dy}=-\frac{x^2-xy+y^2}{y^2}=-(v^2-v+1) \] \[v+y\frac{dv}{dy}=-(v^2-v+1) \] \[y\frac{dv}{dy}=-(v^2+1) \] \[\int\frac{dv}{1+v^2}=-\int\frac{dy}{y} \] \[\tan^{-1}v=-\log|y|+C \] \[\tan^{-1}\frac{x}{y}+\log|y|=C \] Answer: \(\displaystyle \tan^{-1}\dfrac{x}{y}+\log|y|=C\)

  9. Exercise 19

    Solve : (x+y)(dx−dy)=dx+dy\displaystyle (x+y)(d x-d y)=d x+d y. [Hint: Substitute x+y=z\displaystyle x+y=z after seperating dx\displaystyle d x and dy\displaystyle d y ]

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    NCERT’s answer
    \(\displaystyle x+y=k e^{x-y}\)
    Separate \(\displaystyle dx\) and \(\displaystyle dy\): \[(x+y-1)\,dx=(x+y+1)\,dy \] \[\frac{dy}{dx}=\frac{x+y-1}{x+y+1} \] Put \(\displaystyle z=x+y\), so \(\displaystyle \frac{dy}{dx}=\frac{dz}{dx}-1\): \[\frac{dz}{dx}=1+\frac{z-1}{z+1}=\frac{2z}{z+1} \] \[\int\left(1+\frac1z\right)dz=2\int dx \] \[z+\log|z|=2x+C \] \[x+y+\log|x+y|=2x+C \] Answer: \(\displaystyle y-x+\log|x+y|=C\)

  10. Exercise 20

    Solve : 2(y+3)−xydydx=0\displaystyle 2(y+3)-x y \frac{d y}{d x}=0, given that y(1)=−2\displaystyle y(1)=-2.

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    NCERT’s answer
    \(\displaystyle x^2(y+3)^3=e^{y+2}\)
    Separate the variables: \[\frac{y}{y+3}\,dy=\frac{2}{x}\,dx \] \[\int\left(1-\frac{3}{y+3}\right)dy=2\int\frac{dx}{x} \] \[y-3\log|y+3|=2\log|x|+C \] Use \(\displaystyle y(1)=-2\): \[-2-3\log 1=2\log 1+C\;\Rightarrow\;C=-2 \] \[y-3\log|y+3|=2\log|x|-2 \] \[x^2(y+3)^3=e^{\,y+2} \] Answer: \(\displaystyle y-3\log|y+3|=2\log|x|-2\)