NCERT Solutions · Class 12 Mathematics Differential Equations

The equation is linear in \(\displaystyle y\), of the form \(\displaystyle \frac{dy}{dx}+Py=Q\) with \(\displaystyle P=2\), \(\displaystyle Q=\sin x\).\[\mathrm{I.F.}=e^{\int P\,dx}=e^{\int 2\,dx}=e^{2x}\]Multiplying through by the integrating factor makes the left side an exact derivative: \[\frac{d}{dx}\left(y\,e^{2x}\right)=e^{2x}\sin x\qquad\Rightarrow\qquad y\,e^{2x}=\int e^{2x}\sin x\,dx\]Put \(\displaystyle I=\int e^{2x}\sin x\,dx\) and integrate by parts twice: \[I=\frac{e^{2x}}{2}\sin x-\frac{1}{2}\int e^{2x}\cos x\,dx=\frac{e^{2x}}{2}\sin x-\frac{1}{2}\left(\frac{e^{2x}}{2}\cos x+\frac{1}{2}I\right)\] so \(\displaystyle \frac{5}{4}I=\frac{e^{2x}}{4}\left(2\sin x-\cos x\right)\), i.e. \(\displaystyle I=\frac{e^{2x}}{5}\left(2\sin x-\cos x\right)\).Therefore \(\displaystyle y\,e^{2x}=\frac{e^{2x}}{5}\left(2\sin x-\cos x\right)+C\), and dividing by \(\displaystyle e^{2x}\): \[\boxed{\,y=\frac{1}{5}\left(2\sin x-\cos x\right)+C\,e^{-2x}\,}\]

Linear form \(\displaystyle \frac{dy}{dx}+Py=Q\) with \(\displaystyle P=3\), \(\displaystyle Q=e^{-2x}\).\[\mathrm{I.F.}=e^{\int 3\,dx}=e^{3x}\]Multiplying by \(\displaystyle e^{3x}\): \[\frac{d}{dx}\left(y\,e^{3x}\right)=e^{3x}\cdot e^{-2x}=e^{x}\]The step to be careful with is combining the exponentials before integrating — \(\displaystyle e^{3x}e^{-2x}=e^{x}\), not \(\displaystyle e^{-6x^2}\). Integrating, \[y\,e^{3x}=e^{x}+C\]Dividing by \(\displaystyle e^{3x}\): \[\boxed{\,y=e^{-2x}+C\,e^{-3x}\,}\]

Linear form with \(\displaystyle P=\frac{1}{x}\), \(\displaystyle Q=x^{2}\) (so \(\displaystyle x\neq 0\)).\[\mathrm{I.F.}=e^{\int \frac{1}{x}dx}=e^{\log|x|}=x\] (the constant of integration in the exponent is absorbed, and the sign is absorbed into \(\displaystyle C\)).Multiplying by \(\displaystyle x\): \[\frac{d}{dx}\left(xy\right)=x\cdot x^{2}=x^{3}\]Integrating, \[xy=\frac{x^{4}}{4}+C\]Dividing by \(\displaystyle x\): \[\boxed{\,y=\frac{x^{3}}{4}+\frac{C}{x}\,}\]

Linear form with \(\displaystyle P=\sec x\), \(\displaystyle Q=\tan x\), on \(\displaystyle 0\le x<\frac{\pi}{2}\).Using \(\displaystyle \int \sec x\,dx=\log|\sec x+\tan x|\), and noting that on \(\displaystyle 0\le x<\frac{\pi}{2}\) both \(\displaystyle \sec x\ge 1\) and \(\displaystyle \tan x\ge 0\), so \(\displaystyle \sec x+\tan x>0\) and the modulus can be dropped: \[\mathrm{I.F.}=e^{\log(\sec x+\tan x)}=\sec x+\tan x\]Multiplying through: \[\frac{d}{dx}\Big(y(\sec x+\tan x)\Big)=\tan x\,(\sec x+\tan x)=\sec x\tan x+\tan^{2}x\]Now replace \(\displaystyle \tan^{2}x\) by \(\displaystyle \sec^{2}x-1\) before integrating: \[\int\left(\sec x\tan x+\sec^{2}x-1\right)dx=\sec x+\tan x-x+C\]Hence \[\boxed{\,y(\sec x+\tan x)=\sec x+\tan x-x+C\,}\] equivalently \(\displaystyle y=1-\dfrac{x-C}{\sec x+\tan x}\).

First put the equation in standard linear form by dividing by \(\displaystyle \cos^{2}x\) (legitimate on \(\displaystyle 0\le x<\frac{\pi}{2}\), where \(\displaystyle \cos x\neq 0\)): \[\frac{dy}{dx}+\sec^{2}x\;y=\sec^{2}x\,\tan x\]So \(\displaystyle P=\sec^{2}x\) and \[\mathrm{I.F.}=e^{\int \sec^{2}x\,dx}=e^{\tan x}\]Multiplying through: \[\frac{d}{dx}\left(y\,e^{\tan x}\right)=e^{\tan x}\sec^{2}x\,\tan x\qquad\Rightarrow\qquad y\,e^{\tan x}=\int e^{\tan x}\tan x\,\sec^{2}x\,dx\]Substitute \(\displaystyle t=\tan x\), so \(\displaystyle dt=\sec^{2}x\,dx\): \[\int t\,e^{t}\,dt=t\,e^{t}-\int e^{t}dt=(t-1)e^{t}+C=(\tan x-1)e^{\tan x}+C\]Therefore \(\displaystyle y\,e^{\tan x}=(\tan x-1)e^{\tan x}+C\), and dividing by \(\displaystyle e^{\tan x}\): \[\boxed{\,y=(\tan x-1)+C\,e^{-\tan x}\,}\]

Divide by \(\displaystyle x\) (\(\displaystyle x>0\), as \(\displaystyle \log x\) requires) to reach standard linear form: \[\frac{dy}{dx}+\frac{2}{x}y=x\log x\]So \(\displaystyle P=\frac{2}{x}\) and \[\mathrm{I.F.}=e^{\int \frac{2}{x}dx}=e^{2\log x}=e^{\log x^{2}}=x^{2}\]Multiplying through: \[\frac{d}{dx}\left(x^{2}y\right)=x^{2}\cdot x\log x=x^{3}\log x\]Integrate by parts taking \(\displaystyle \log x\) as the first function: \[\int x^{3}\log x\,dx=\log x\cdot\frac{x^{4}}{4}-\int\frac{1}{x}\cdot\frac{x^{4}}{4}\,dx=\frac{x^{4}}{4}\log x-\frac{x^{4}}{16}+C\]Hence \(\displaystyle x^{2}y=\frac{x^{4}}{4}\log x-\frac{x^{4}}{16}+C\), and dividing by \(\displaystyle x^{2}\): \[\boxed{\,y=\frac{x^{2}}{4}\log x-\frac{x^{2}}{16}+\frac{C}{x^{2}}\,}\]

Divide by \(\displaystyle x\log x\) (so \(\displaystyle x>0\) and \(\displaystyle x\neq 1\)) to get standard linear form: \[\frac{dy}{dx}+\frac{1}{x\log x}\,y=\frac{2}{x^{2}}\]For the integrating factor, substitute \(\displaystyle t=\log x\), \(\displaystyle dt=\frac{dx}{x}\): \[\int\frac{dx}{x\log x}=\int\frac{dt}{t}=\log|t|=\log|\log x|\quad\Rightarrow\quad \mathrm{I.F.}=e^{\log|\log x|}=\log x\]Multiplying through: \[\frac{d}{dx}\Big(y\log x\Big)=\frac{2\log x}{x^{2}}\]Integrate the right side by parts with \(\displaystyle \log x\) first and \(\displaystyle \frac{2}{x^{2}}\) second: \[\int \frac{2\log x}{x^{2}}dx=\log x\left(-\frac{2}{x}\right)-\int\frac{1}{x}\left(-\frac{2}{x}\right)dx=-\frac{2\log x}{x}+2\int\frac{dx}{x^{2}}=-\frac{2\log x}{x}-\frac{2}{x}+C\]Therefore \[\boxed{\,y\log x=-\frac{2}{x}\left(1+\log x\right)+C\,}\]

Write the equation with \(\displaystyle x\) as the independent variable: \[\left(1+x^{2}\right)\frac{dy}{dx}+2xy=\cot x\qquad\Rightarrow\qquad \frac{dy}{dx}+\frac{2x}{1+x^{2}}\,y=\frac{\cot x}{1+x^{2}}\]This is linear with \(\displaystyle P=\frac{2x}{1+x^{2}}\): \[\int\frac{2x}{1+x^{2}}dx=\log\left(1+x^{2}\right)\quad\Rightarrow\quad \mathrm{I.F.}=e^{\log(1+x^{2})}=1+x^{2}\](Notice that the original left side, \(\displaystyle \left(1+x^{2}\right)dy+2xy\,dx\), is already \(\displaystyle d\!\left[(1+x^{2})y\right]\) — the integrating factor is \(\displaystyle 1\) times what is there.)Multiplying by \(\displaystyle 1+x^{2}\): \[\frac{d}{dx}\Big(\left(1+x^{2}\right)y\Big)=\cot x\qquad\Rightarrow\qquad \left(1+x^{2}\right)y=\int\cot x\,dx=\log|\sin x|+C\]Hence \[\boxed{\,y\left(1+x^{2}\right)=\log|\sin x|+C\,}\] valid where \(\displaystyle \sin x\neq 0\).

Divide by \(\displaystyle x\) (\(\displaystyle x\neq 0\)) and collect the terms in \(\displaystyle y\): \[\frac{dy}{dx}+\frac{y}{x}-1+y\cot x=0\qquad\Rightarrow\qquad \frac{dy}{dx}+\left(\frac{1}{x}+\cot x\right)y=1\]This is linear with \(\displaystyle P=\frac{1}{x}+\cot x\): \[\int P\,dx=\log|x|+\log|\sin x|=\log|x\sin x|\quad\Rightarrow\quad \mathrm{I.F.}=x\sin x\]Multiplying through: \[\frac{d}{dx}\Big(y\,x\sin x\Big)=x\sin x\]Integrate the right side by parts (\(\displaystyle x\) first, \(\displaystyle \sin x\) second): \[\int x\sin x\,dx=-x\cos x+\int\cos x\,dx=-x\cos x+\sin x+C\]So \(\displaystyle y\,x\sin x=-x\cos x+\sin x+C\). Dividing by \(\displaystyle x\sin x\): \[\boxed{\,y=-\cot x+\frac{1}{x}+\frac{C}{x\sin x}\,}\]

As written, \(\displaystyle \frac{dy}{dx}=\frac{1}{x+y}\) is not linear in \(\displaystyle y\). The key step is to invert the derivative and treat \(\displaystyle x\) as the dependent variable and \(\displaystyle y\) as the independent one: \[\frac{dx}{dy}=x+y\qquad\Rightarrow\qquad \frac{dx}{dy}-x=y\]This is linear in \(\displaystyle x\) with \(\displaystyle P_{1}=-1\), \(\displaystyle Q_{1}=y\): \[\mathrm{I.F.}=e^{\int(-1)dy}=e^{-y}\]Multiplying through: \[\frac{d}{dy}\left(x\,e^{-y}\right)=y\,e^{-y}\]Integrating by parts, \[\int y\,e^{-y}dy=-y\,e^{-y}+\int e^{-y}dy=-y\,e^{-y}-e^{-y}+C=-(y+1)e^{-y}+C\]So \(\displaystyle x\,e^{-y}=-(y+1)e^{-y}+C\). Multiplying by \(\displaystyle e^{y}\): \[\boxed{\,x+y+1=C\,e^{y}\,}\]