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NCERT Solutions · Class 12 Mathematics Differential Equations

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EXERCISE 9.3 1–10 (part 3 of 10)

  1. For each of the differential equations in Exercises $\displaystyle 1$ to $\displaystyle 10$, find the general solution:

    Exercise 1

    dydx=1−cos⁡x1+cos⁡x\displaystyle \frac{d y}{d x}=\frac{1-\cos x}{1+\cos x}

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    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 \(\displaystyle 1-\cos x = 2\sin^{2}\frac{x}{2}\) and \(\displaystyle 1+\cos x = 2\cos^{2}\frac{x}{2}\): \[\frac{dy}{dx}=\frac{2\sin^{2}\dfrac{x}{2}}{2\cos^{2}\dfrac{x}{2}}=\tan^{2}\frac{x}{2}\] The step most often missed: \(\displaystyle \tan^{2}\theta\) has no direct antiderivative, so convert it by \(\displaystyle \tan^{2}\theta=\sec^{2}\theta-1\): \[dy=\left(\sec^{2}\frac{x}{2}-1\right)dx\] Integrating both sides, and remembering that \(\displaystyle \int \sec^{2}\frac{x}{2}\,dx = 2\tan\frac{x}{2}\) because of the inner factor \(\displaystyle \tfrac12\), \[y = 2\tan\frac{x}{2}-x+C\] General solution: \(\displaystyle y = 2\tan\dfrac{x}{2} - x + C\), valid wherever \(\displaystyle \cos x \neq -1\).

  2. Exercise 2

    dydx=4−y2(−2<y<2)\displaystyle \frac{d y}{d x}=\sqrt{4-y^{2}} \quad(-2<y<2)

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    NCERT’s answer
    \(\displaystyle y=2 \sin (x+\mathrm{C})\)
    Variables separable. For \(\displaystyle -2<y<2\) we have \(\displaystyle 4-y^{2}>0\), so we may divide by \(\displaystyle \sqrt{4-y^{2}}\): \[\frac{dy}{\sqrt{4-y^{2}}}=dx\] Use the standard integral \(\displaystyle \int \frac{dy}{\sqrt{a^{2}-y^{2}}}=\sin^{-1}\frac{y}{a}+C\) with \(\displaystyle a=2\); the restriction \(\displaystyle -2<y<2\) is exactly what makes \(\displaystyle \frac{y}{2}\) lie in \(\displaystyle (-1,1)\), the domain of \(\displaystyle \sin^{-1}\): \[\sin^{-1}\frac{y}{2}=x+C\] Solving for \(\displaystyle y\), \[\frac{y}{2}=\sin(x+C)\] General solution: \(\displaystyle y = 2\sin(x+C)\), with \(\displaystyle C\) arbitrary.

  3. Exercise 3

    dydx+y=1(y≠1)\displaystyle \frac{d y}{d x}+y=1(y \neq 1)

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    Write the equation as \(\displaystyle \dfrac{dy}{dx}=1-y\). Since \(\displaystyle y\neq 1\) we may divide by \(\displaystyle 1-y\) and separate: \[\frac{dy}{1-y}=dx\] Integrating, and noting \(\displaystyle \int \frac{dy}{1-y} = -\log|1-y|\) (the minus sign from the inner derivative is the usual slip), \[-\log|1-y| = x + C_{1}\] \[\log|1-y| = -x - C_{1}\] \[|1-y| = e^{-C_{1}}e^{-x}\quad\Rightarrow\quad 1-y = A e^{-x},\ A = \pm e^{-C_{1}}\neq 0\] Hence \(\displaystyle y = 1 - Ae^{-x}\). Writing \(\displaystyle C=-A\), \[y = 1 + Ce^{-x}\] General solution: \(\displaystyle y = 1 + Ce^{-x}\).

  4. Exercise 4

    sec⁡2xtan⁡ydx+sec⁡2ytan⁡xdy=0\displaystyle \sec ^{2} x \tan y d x+\sec ^{2} y \tan x d y=0

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    NCERT’s answer
    \(\displaystyle \tan x \tan y=\mathrm{C}\)
    The equation is variables separable once we divide throughout by \(\displaystyle \tan x\,\tan y\) (permissible where \(\displaystyle \tan x \neq 0,\ \tan y \neq 0\)): \[\frac{\sec^{2}x}{\tan x}\,dx + \frac{\sec^{2}y}{\tan y}\,dy = 0\] Each integral is of the form \(\displaystyle \int \frac{f'(t)}{f(t)}dt = \log|f(t)|\), since \(\displaystyle \frac{d}{dx}(\tan x)=\sec^{2}x\): \[\log|\tan x| + \log|\tan y| = \log C\] \[\log|\tan x \tan y| = \log C\] General solution: \(\displaystyle \tan x\,\tan y = C\).

  5. Exercise 5

    (ex+e−x)dy−(ex−e−x)dx=0\displaystyle \left(e^{x}+e^{-x}\right) d y-\left(e^{x}-e^{-x}\right) d x=0

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    Rearranging, \[\frac{dy}{dx}=\frac{e^{x}-e^{-x}}{e^{x}+e^{-x}}\] Observe that the numerator is exactly the derivative of the denominator: \(\displaystyle \frac{d}{dx}\left(e^{x}+e^{-x}\right)=e^{x}-e^{-x}\). So this is again \(\displaystyle \int \frac{f'(x)}{f(x)}dx\): \[y=\int \frac{e^{x}-e^{-x}}{e^{x}+e^{-x}}\,dx = \log\left(e^{x}+e^{-x}\right)+C\] No modulus sign is needed because \(\displaystyle e^{x}+e^{-x}>0\) for every real \(\displaystyle x\). General solution: \(\displaystyle y = \log\left(e^{x}+e^{-x}\right)+C\).

  6. Exercise 6

    dydx=(1+x2)(1+y2)\displaystyle \frac{d y}{d x}=\left(1+x^{2}\right)\left(1+y^{2}\right)

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    NCERT’s answer
    \(\displaystyle \tan ^{-1} y=x+\frac{x^{3}}{3}+\mathrm{C}\)
    The right-hand side already factors as a function of \(\displaystyle x\) times a function of \(\displaystyle y\), so separate: \[\frac{dy}{1+y^{2}}=\left(1+x^{2}\right)dx\] Integrate, using \(\displaystyle \int \frac{dy}{1+y^{2}}=\tan^{-1}y\): \[\tan^{-1}y = x + \frac{x^{3}}{3}+C\] General solution: \(\displaystyle \tan^{-1}y = x + \dfrac{x^{3}}{3}+C\), i.e. \(\displaystyle y = \tan\left(x+\dfrac{x^{3}}{3}+C\right)\).

  7. Exercise 7

    ylog⁡ydx−xdy=0\displaystyle y \log y d x-x d y=0

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    Separate the variables (this needs \(\displaystyle x\neq 0\), \(\displaystyle y>0\) and \(\displaystyle \log y \neq 0\), i.e. \(\displaystyle y\neq 1\)): \[y\log y\,dx = x\,dy \quad\Rightarrow\quad \frac{dx}{x}=\frac{dy}{y\log y}\] For the right side substitute \(\displaystyle t=\log y\), so \(\displaystyle dt = \dfrac{dy}{y}\): \[\int \frac{dy}{y\log y}=\int \frac{dt}{t}=\log|t| = \log|\log y|\] Therefore \[\log|\log y| = \log|x| + \log C\] \[\log y = Cx\] General solution: \(\displaystyle y = e^{Cx}\).

  8. Exercise 8

    x5dydx=−y5\displaystyle x^{5} \frac{d y}{d x}=-y^{5}

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    NCERT’s answer
    \(\displaystyle x^{-4}+y^{-4}=\mathrm{C}\)
    Separate the variables (valid for \(\displaystyle x\neq 0,\ y\neq 0\)): \[\frac{dy}{y^{5}}=-\frac{dx}{x^{5}}\] \[\int y^{-5}\,dy = -\int x^{-5}\,dx\] Watch the two negative signs here, in the power rule and in the equation itself: \[\frac{y^{-4}}{-4}= -\frac{x^{-4}}{-4}+C_{1}\quad\Rightarrow\quad -\frac{1}{4y^{4}}=\frac{1}{4x^{4}}+C_{1}\] Multiplying through by \(\displaystyle -4\) and writing \(\displaystyle C=-4C_{1}\): \[\frac{1}{x^{4}}+\frac{1}{y^{4}}=C\] General solution: \(\displaystyle x^{-4}+y^{-4}=C\).

  9. Exercise 9

    dydx=sin⁡−1x\displaystyle \frac{d y}{d x}=\sin ^{-1} x

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    NCERT’s answer
    \(\displaystyle y=x \sin ^{-1} x+\sqrt{1-x^{2}}+\mathrm{C}\)
    Direct integration: \(\displaystyle y=\int \sin^{-1}x\,dx\). Integrate by parts with \(\displaystyle u=\sin^{-1}x\), \(\displaystyle dv = dx\) (the standard trick of taking the second function to be \(\displaystyle 1\)): \[\int \sin^{-1}x\cdot 1\,dx = x\sin^{-1}x - \int \frac{x}{\sqrt{1-x^{2}}}\,dx\] For the remaining integral put \(\displaystyle t = 1-x^{2}\), \(\displaystyle dt = -2x\,dx\): \[\int \frac{x}{\sqrt{1-x^{2}}}\,dx = -\frac{1}{2}\int \frac{dt}{\sqrt{t}} = -\sqrt{1-x^{2}}\] Hence \[y = x\sin^{-1}x + \sqrt{1-x^{2}}+C\] General solution: \(\displaystyle y = x\sin^{-1}x + \sqrt{1-x^{2}}+C\), for \(\displaystyle -1\le x\le 1\).

  10. Exercise 10

    extan⁡ydx+(1−ex)sec⁡2ydy=0\displaystyle e^{x} \tan y d x+\left(1-e^{x}\right) \sec ^{2} y d y=0

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    NCERT’s answer
    \(\displaystyle \tan y=\mathrm{C}\left(1-e^{x}\right)\)
    Divide throughout by \(\displaystyle \left(1-e^{x}\right)\tan y\) to separate (needs \(\displaystyle e^{x}\neq 1\), \(\displaystyle \tan y \neq 0\)): \[\frac{e^{x}}{1-e^{x}}\,dx + \frac{\sec^{2}y}{\tan y}\,dy = 0\] In the first integral the numerator is \(\displaystyle -1\) times the derivative of the denominator, since \(\displaystyle \frac{d}{dx}\left(1-e^{x}\right)=-e^{x}\); in the second, \(\displaystyle \frac{d}{dy}(\tan y)=\sec^{2}y\): \[-\log\left|1-e^{x}\right| + \log|\tan y| = \log C\] \[\log\left|\frac{\tan y}{1-e^{x}}\right| = \log C\] General solution: \(\displaystyle \tan y = C\left(1-e^{x}\right)\).