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NCERT Exemplar · Class 12 Mathematics Differential Equations

77 questions · 77 still being checked

EXERCISE 9.3 21–30 (part 3 of 8)

  1. Exercise 21

    Solve the differential equation dy=cos⁡x(2−ycosec⁡x)dx\displaystyle d y=\cos x(2-y \operatorname{cosec} x) d x given that y=2\displaystyle y=2 when x=π2.x=\frac{\pi}{2} .

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    NCERT’s answer
    \(\displaystyle y \sin x=\frac{-\cos 2 x}{2}+\frac{3}{2}\)
    Rewrite as a linear equation: \[\frac{dy}{dx}=2\cos x-y\cot x \] \[\frac{dy}{dx}+y\cot x=2\cos x \] \[\text{I.F.}=e^{\int\cot x\,dx}=\sin x \] \[y\sin x=\int2\sin x\cos x\,dx=\sin^2x+C \] Use \(\displaystyle y=2\) at \(\displaystyle x=\frac{\pi}{2}\): \[2\cdot1=1+C\;\Rightarrow\;C=1 \] \[y\sin x=\sin^2x+1 \] Answer: \(\displaystyle y=\sin x+\operatorname{cosec}x\)

  2. Exercise 22

    Form the differential equation by eliminating A and B in Ax2+By2=1\displaystyle \mathrm{A} x^2+\mathrm{B} y^2=1.

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    Differentiate the relation twice with respect to \(\displaystyle x\): \[Ax^2+By^2=1 \] \[Ax+By\,y'=0 \] \[A+B\left(y'^2+y\,y''\right)=0 \] From the first derivative \(\displaystyle A=-\dfrac{By\,y'}{x}\); substitute in the second: \[-\frac{By\,y'}{x}+B\left(y'^2+y\,y''\right)=0 \] \[xy\,y''+x\,y'^2-y\,y'=0 \] Answer: \(\displaystyle xy\dfrac{d^2y}{dx^2}+x\left(\dfrac{dy}{dx}\right)^2-y\dfrac{dy}{dx}=0\)

  3. Exercise 23

    Solve the differential equation (1+y2)tan⁡−1xdx+2y(1+x2)dy=0\displaystyle \left(1+y^2\right) \tan ^{-1} x d x+2 y\left(1+x^2\right) d y=0.

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    NCERT’s answer
    \(\displaystyle \frac{1}{2}\left(\tan ^{-1} x\right)^2+\log \left(1+y^2\right)=c\)
    Divide by \(\displaystyle (1+x^2)(1+y^2)\): \[\frac{\tan^{-1}x}{1+x^2}\,dx+\frac{2y}{1+y^2}\,dy=0 \] \[\int\frac{\tan^{-1}x}{1+x^2}\,dx+\int\frac{2y}{1+y^2}\,dy=C_1 \] \[\frac{(\tan^{-1}x)^2}{2}+\log(1+y^2)=C_1 \] Answer: \(\displaystyle (\tan^{-1}x)^2+2\log(1+y^2)=C\)

  4. Exercise 24

    Find the differential equation of system of concentric circles with centre (1,2)\displaystyle (1,2).

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    NCERT’s answer
    \(\displaystyle (x-1)+(y-2) \frac{d y}{d x}=0\)
    NCERT_Solution_Class12_Maths_Exemplar_Ch9_Ex9-3_Q24 Circles with centre \(\displaystyle (1,2)\) and radius \(\displaystyle r\): \[(x-1)^2+(y-2)^2=r^2 \] Differentiate to eliminate \(\displaystyle r\): \[2(x-1)+2(y-2)\frac{dy}{dx}=0 \] \[(x-1)+(y-2)\frac{dy}{dx}=0 \] Answer: \(\displaystyle (x-1)+(y-2)\dfrac{dy}{dx}=0\)

  5. Exercise 25

    Solve : y+ddx(xy)=x(sin⁡x+log⁡x)\displaystyle y+\frac{d}{d x}(x y)=x(\sin x+\log x)

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    NCERT’s answer
    \(\displaystyle y=-\cos x+\frac{2 \sin x}{x}+\frac{2 \cos x}{x^2}+\frac{x \log x}{3}-\frac{x}{9}+c x^{-2}\)
    Expand the product: \[y + y + x\frac{dy}{dx} = x(\sin x + \log x) \] \[\frac{dy}{dx} + \frac{2}{x}\,y = \sin x + \log x \] Linear in \(\displaystyle y\): \[\text{I.F.} = e^{\int \frac{2}{x}\,dx} = x^2 \] \[\frac{d}{dx}\left(x^2 y\right) = x^2\sin x + x^2\log x \] By parts: \[\int x^2\sin x\,dx = -x^2\cos x + 2\int x\cos x\,dx = -x^2\cos x + 2x\sin x + 2\cos x \] \[\int x^2\log x\,dx = \frac{x^3}{3}\log x - \int\frac{x^2}{3}\,dx = \frac{x^3}{3}\log x - \frac{x^3}{9} \] \[x^2 y = -x^2\cos x + 2x\sin x + 2\cos x + \frac{x^3}{3}\log x - \frac{x^3}{9} + C \] Answer: \(\displaystyle y = -\cos x + \dfrac{2\sin x}{x} + \dfrac{2\cos x}{x^2} + \dfrac{x}{3}\log x - \dfrac{x}{9} + \dfrac{C}{x^2} \)

  6. Exercise 26

    Find the general solution of (1+tan⁡y)(dx−dy)+2xdy=0\displaystyle (1+\tan y)(d x-d y)+2 x d y=0.

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    NCERT’s answer
    \(\displaystyle x(\sin y+\cos y)=\sin y+c e^{-y}\)
    Take \(\displaystyle x\) as the dependent variable: \[(1+\tan y)\,dx + (2x - 1 - \tan y)\,dy = 0 \] \[\frac{dx}{dy} + \frac{2}{1+\tan y}\,x = 1 \] \[\frac{2}{1+\tan y} = \frac{2\cos y}{\cos y+\sin y} = 1 + \frac{\cos y - \sin y}{\cos y + \sin y} \] \[\int\frac{2}{1+\tan y}\,dy = y + \log(\sin y + \cos y) \] \[\text{I.F.} = e^{y}(\sin y + \cos y) \] \[x\,e^{y}(\sin y + \cos y) = \int e^{y}(\sin y + \cos y)\,dy \] Since \[\frac{d}{dy}\left(e^{y}\sin y\right) = e^{y}(\sin y + \cos y) \] \[x\,e^{y}(\sin y + \cos y) = e^{y}\sin y + C \] Answer: \(\displaystyle x(\sin y + \cos y) = \sin y + Ce^{-y} \)

  7. Exercise 27

    Solve : dydx=cos⁡(x+y)+sin⁡(x+y)\displaystyle \frac{d y}{d x}=\cos (x+y)+\sin (x+y).[Hint: Substitute x+y=z\displaystyle x+y=z ]

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    NCERT’s answer
    \(\displaystyle \log \left|1+\tan \left(\frac{x+y}{2}\right)\right|=x+c\)
    Put \(\displaystyle z = x + y\): \[\frac{dz}{dx} = 1 + \cos z + \sin z \] \[1 + \cos z + \sin z = 2\cos^2\frac z2 + 2\sin\frac z2\cos\frac z2 = 2\cos^2\frac z2\left(1 + \tan\frac z2\right) \] \[\frac{\tfrac12\sec^2\frac z2\,dz}{1 + \tan\frac z2} = dx \] \[\log\left|1 + \tan\frac z2\right| = x + c \] \[1 + \tan\frac{x+y}{2} = Ce^{x} \] Answer: \(\displaystyle \tan\dfrac{x+y}{2} = Ce^{x} - 1 \)

  8. Exercise 28

    Find the general solution of dydx−3y=sin⁡2x\displaystyle \frac{d y}{d x}-3 y=\sin 2 x.

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    NCERT’s answer
    \(\displaystyle y=-\left[\frac{3 \sin 2 x+2 \cos 2 x}{13}\right]+c e^{3 x}\)
    Linear in \(\displaystyle y\): \[\text{I.F.} = e^{\int -3\,dx} = e^{-3x} \] \[\frac{d}{dx}\left(y\,e^{-3x}\right) = e^{-3x}\sin 2x \] \[\int e^{ax}\sin bx\,dx = \frac{e^{ax}(a\sin bx - b\cos bx)}{a^2+b^2} \] With \(\displaystyle a=-3,\ b=2\): \[y\,e^{-3x} = \frac{e^{-3x}\left(-3\sin 2x - 2\cos 2x\right)}{13} + C \] Answer: \(\displaystyle y = -\dfrac{3\sin 2x + 2\cos 2x}{13} + Ce^{3x} \)

  9. Exercise 29

    Find the equation of a curve passing through (2,1)\displaystyle (2,1) if the slope of the tangent to the curve at any point (x,y)\displaystyle (x, y) is x2+y22xy\displaystyle \frac{x^2+y^2}{2 x y}.

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    NCERT’s answer
    \(\displaystyle 2\left(x^2-y^2\right)=3 x\)
    Slope of the tangent: \[\frac{dy}{dx} = \frac{x^2+y^2}{2xy} \] Homogeneous; put \(\displaystyle y = vx\): \[v + x\frac{dv}{dx} = \frac{1+v^2}{2v} \] \[x\frac{dv}{dx} = \frac{1-v^2}{2v} \] \[\frac{2v\,dv}{1-v^2} = \frac{dx}{x} \] \[-\log|1-v^2| = \log|x| + \log|c| \] \[x(1-v^2) = C \] \[x^2 - y^2 = Cx \] Through \(\displaystyle (2,1)\): \[4 - 1 = 2C \Rightarrow C = \frac32 \] Answer: \(\displaystyle x^2 - y^2 = \dfrac32 x \), i.e. \(\displaystyle 2x^2 - 2y^2 = 3x \)

  10. Exercise 30

    Find the equation of the curve through the point (1,0)\displaystyle (1,0) if the slope of the tangent to the curve at any point (x,y)\displaystyle (x, y) is y−1x2+x\displaystyle \frac{y-1}{x^2+x}.

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    NCERT’s answer
    \(\displaystyle (y-1)(x+1)+2 x=0\)
    Slope of the tangent: \[\frac{dy}{dx} = \frac{y-1}{x^2+x} \] \[\frac{dy}{y-1} = \frac{dx}{x(x+1)} = \left(\frac1x - \frac1{x+1}\right)dx \] \[\log|y-1| = \log\left|\frac{x}{x+1}\right| + \log|C| \] \[y - 1 = \frac{Cx}{x+1} \] Through \(\displaystyle (1,0)\): \[-1 = \frac{C}{2} \Rightarrow C = -2 \] \[y = 1 - \frac{2x}{x+1} = \frac{1-x}{1+x} \] Answer: \(\displaystyle y = \dfrac{1-x}{1+x} \), i.e. \(\displaystyle xy + x + y - 1 = 0 \)