NCERT Solutions · Class 12 Mathematics Differential Equations

The equation \(\displaystyle \dfrac{dy}{dx}+y\cot x=4x\,\mathrm{cosec}\,x\) is linear in \(\displaystyle y\), with \(\displaystyle \mathrm{P}=\cot x\) and \(\displaystyle \mathrm{Q}=4x\,\mathrm{cosec}\,x\).Integrating factor: \[\int \mathrm{P}\,dx=\int\cot x\,dx=\log|\sin x|,\qquad \text{I.F.}=e^{\log|\sin x|}=\sin x.\]Then \(\displaystyle y\cdot(\text{I.F.})=\int \mathrm{Q}\cdot(\text{I.F.})\,dx+\mathrm{C}\), and here \(\displaystyle \mathrm{cosec}\,x\cdot\sin x=1\), which is the simplification the question is built on: \[y\sin x=\int 4x\,\mathrm{cosec}\,x\cdot\sin x\,dx+\mathrm{C}=\int 4x\,dx+\mathrm{C}=2x^{2}+\mathrm{C}.\]Apply \(\displaystyle y=0\) when \(\displaystyle x=\dfrac{\pi}{2}\), where \(\displaystyle \sin x=1\): \[0=2\cdot\frac{\pi^{2}}{4}+\mathrm{C}\quad\Longrightarrow\quad \mathrm{C}=-\frac{\pi^{2}}{2}.\]Particular solution: \(\displaystyle y\sin x=2x^{2}-\dfrac{\pi^{2}}{2}\), i.e. \(\displaystyle y=\dfrac{2x^{2}-\frac{\pi^{2}}{2}}{\sin x}\), \(\displaystyle \sin x\neq 0\).

The equation \(\displaystyle (x+1)\dfrac{dy}{dx}=2e^{-y}-1\) has the variables separable: \[\frac{dy}{2e^{-y}-1}=\frac{dx}{x+1}.\]The left side is awkward as it stands; multiply numerator and denominator by \(\displaystyle e^{y}\): \[\frac{e^{y}\,dy}{2-e^{y}}=\frac{dx}{x+1}.\] For the left integral put \(\displaystyle u=2-e^{y}\), so \(\displaystyle du=-e^{y}dy\) and \(\displaystyle \int\frac{e^{y}dy}{2-e^{y}}=-\log|2-e^{y}|\). Hence \[-\log\left|2-e^{y}\right|=\log|x+1|+\log \mathrm{C},\] which rearranges to \[\log\frac{1}{\left|2-e^{y}\right|}=\log\big(\mathrm{C}|x+1|\big)\quad\Longrightarrow\quad \frac{1}{2-e^{y}}=\mathrm{C}\,(x+1).\]Apply \(\displaystyle y=0\) when \(\displaystyle x=0\): \[\frac{1}{2-e^{0}}=\mathrm{C}(0+1)\quad\Longrightarrow\quad \frac{1}{1}=\mathrm{C}\quad\Longrightarrow\quad \mathrm{C}=1.\]So \(\displaystyle \dfrac{1}{2-e^{y}}=x+1\), giving \(\displaystyle 2-e^{y}=\dfrac{1}{x+1}\) and \[e^{y}=2-\frac{1}{x+1}=\frac{2x+1}{x+1}.\]Particular solution: \(\displaystyle y=\log\left|\dfrac{2x+1}{x+1}\right|\), \(\displaystyle x\neq-1\).

Since \(\displaystyle \dfrac{y\,dx-x\,dy}{y}=0\) with \(\displaystyle y\neq 0\), the numerator must vanish: \[y\,dx-x\,dy=0\quad\Longrightarrow\quad \frac{dy}{y}=\frac{dx}{x}.\]Integrating both sides, \[\log|y|=\log|x|+\log \mathrm{C}\quad\Longrightarrow\quad y=\mathrm{C}x.\]Answer: (C) \(\displaystyle y=\mathrm{C}x\).

In \(\displaystyle \dfrac{dx}{dy}+\mathrm{P}_{1}x=\mathrm{Q}_{1}\) the roles are interchanged: \(\displaystyle x\) is the dependent variable and \(\displaystyle y\) the independent one, with \(\displaystyle \mathrm{P}_{1},\mathrm{Q}_{1}\) functions of \(\displaystyle y\). So every \(\displaystyle dx\) in the usual recipe becomes \(\displaystyle dy\), and \(\displaystyle y\) in the usual recipe becomes \(\displaystyle x\).Integrating factor: \(\displaystyle \text{I.F.}=e^{\int \mathrm{P}_{1}dy}\). Multiplying the equation by it makes the left side \(\displaystyle \dfrac{d}{dy}\left(x\,e^{\int \mathrm{P}_{1}dy}\right)\), and integrating with respect to \(\displaystyle y\) gives \[x\,e^{\int \mathrm{P}_{1}dy}=\int\left(\mathrm{Q}_{1}e^{\int \mathrm{P}_{1}dy}\right)dy+\mathrm{C}.\]Answer: (C).

Divide \(\displaystyle e^{x}dy+\left(ye^{x}+2x\right)dx=0\) by \(\displaystyle e^{x}\,dx\): \[\frac{dy}{dx}+y=-2x\,e^{-x},\] a linear equation with \(\displaystyle \mathrm{P}=1\) and \(\displaystyle \mathrm{Q}=-2xe^{-x}\).Integrating factor: \(\displaystyle \text{I.F.}=e^{\int 1\,dx}=e^{x}\). Then \[y\,e^{x}=\int e^{x}\left(-2xe^{-x}\right)dx+\mathrm{C}=\int(-2x)\,dx+\mathrm{C}=-x^{2}+\mathrm{C},\] the exponentials cancelling. Hence \[y\,e^{x}+x^{2}=\mathrm{C}.\](The same result follows by noticing the equation is exact: \(\displaystyle e^{x}dy+ye^{x}dx=d\left(ye^{x}\right)\) and \(\displaystyle 2x\,dx=d\left(x^{2}\right)\), so \(\displaystyle d\left(ye^{x}+x^{2}\right)=0\).)Answer: (C) \(\displaystyle ye^{x}+x^{2}=\mathrm{C}\).