NCERT Solutions · Class 12 Mathematics Differential Equations

Rewrite the equation as \[y\frac{dx}{dy}+\left(x-y^{2}\right)=0\qquad\Rightarrow\qquad y\frac{dx}{dy}+x=y^{2}\]It is not linear in \(\displaystyle y\), but it is linear in \(\displaystyle x\). Dividing by \(\displaystyle y\) (\(\displaystyle y\neq 0\)): \[\frac{dx}{dy}+\frac{1}{y}\,x=y\]Hence \[\mathrm{I.F.}=e^{\int\frac{1}{y}dy}=e^{\log|y|}=y\]Multiplying through: \[\frac{d}{dy}\left(xy\right)=y\cdot y=y^{2}\qquad\Rightarrow\qquad xy=\frac{y^{3}}{3}+C\]Therefore \[\boxed{\,xy=\frac{y^{3}}{3}+C\,}\qquad\text{i.e.}\qquad x=\frac{y^{2}}{3}+\frac{C}{y}\]

Here \(\displaystyle \frac{dy}{dx}=\frac{y}{x+3y^{2}}\) is not linear in \(\displaystyle y\), so take \(\displaystyle y\) as the independent variable: \[\frac{dx}{dy}=\frac{x+3y^{2}}{y}=\frac{x}{y}+3y\qquad\Rightarrow\qquad \frac{dx}{dy}-\frac{1}{y}\,x=3y\]This is linear in \(\displaystyle x\) with \(\displaystyle P_{1}=-\frac{1}{y}\). Since \(\displaystyle y>0\), \[\mathrm{I.F.}=e^{-\int\frac{1}{y}dy}=e^{-\log y}=\frac{1}{y}\]Multiplying through: \[\frac{d}{dy}\left(\frac{x}{y}\right)=3y\cdot\frac{1}{y}=3\qquad\Rightarrow\qquad \frac{x}{y}=3y+C\]Therefore \[\boxed{\,x=3y^{2}+Cy\,}\qquad (y>0)\]

Linear form with \(\displaystyle P=2\tan x\), \(\displaystyle Q=\sin x\).\[\int 2\tan x\,dx=2\log|\sec x|=\log\sec^{2}x\quad\Rightarrow\quad \mathrm{I.F.}=e^{\log\sec^{2}x}=\sec^{2}x\]Multiplying through: \[\frac{d}{dx}\left(y\sec^{2}x\right)=\sin x\,\sec^{2}x=\sec x\tan x\]That rewriting is the step to get right: \(\displaystyle \sin x\sec^{2}x=\frac{\sin x}{\cos^{2}x}=\sec x\tan x\). Integrating, \[y\sec^{2}x=\sec x+C\qquad\Rightarrow\qquad y=\cos x+C\cos^{2}x\]Apply \(\displaystyle y=0\) when \(\displaystyle x=\frac{\pi}{3}\), where \(\displaystyle \cos\frac{\pi}{3}=\frac{1}{2}\): \[0=\frac{1}{2}+C\cdot\frac{1}{4}\qquad\Rightarrow\qquad C=-2\]The particular solution is \[\boxed{\,y=\cos x-2\cos^{2}x\,}\]

Divide by \(\displaystyle 1+x^{2}\) to reach standard linear form: \[\frac{dy}{dx}+\frac{2x}{1+x^{2}}\,y=\frac{1}{\left(1+x^{2}\right)^{2}}\]Note that dividing squares the denominator on the right — that is the step most often dropped. Now \[\mathrm{I.F.}=e^{\int\frac{2x}{1+x^{2}}dx}=e^{\log(1+x^{2})}=1+x^{2}\]Multiplying through: \[\frac{d}{dx}\Big(\left(1+x^{2}\right)y\Big)=\frac{1}{1+x^{2}}\qquad\Rightarrow\qquad \left(1+x^{2}\right)y=\tan^{-1}x+C\]Apply \(\displaystyle y=0\) when \(\displaystyle x=1\), where \(\displaystyle \tan^{-1}1=\frac{\pi}{4}\): \[0=\frac{\pi}{4}+C\qquad\Rightarrow\qquad C=-\frac{\pi}{4}\]The particular solution is \[\boxed{\,y\left(1+x^{2}\right)=\tan^{-1}x-\frac{\pi}{4}\,}\]

Linear form with \(\displaystyle P=-3\cot x\), \(\displaystyle Q=\sin 2x\).\[\int(-3\cot x)\,dx=-3\log|\sin x|=\log\left|\sin x\right|^{-3}\quad\Rightarrow\quad \mathrm{I.F.}=\frac{1}{\sin^{3}x}=\mathrm{cosec}^{3}x\]Multiplying through: \[\frac{d}{dx}\left(y\,\mathrm{cosec}^{3}x\right)=\sin 2x\;\mathrm{cosec}^{3}x=\frac{2\sin x\cos x}{\sin^{3}x}=\frac{2\cos x}{\sin^{2}x}\]Using \(\displaystyle \sin 2x=2\sin x\cos x\) is what makes this integrable. With \(\displaystyle t=\sin x\), \(\displaystyle dt=\cos x\,dx\): \[\int\frac{2\cos x}{\sin^{2}x}dx=2\int\frac{dt}{t^{2}}=-\frac{2}{t}=-\frac{2}{\sin x}\]So \(\displaystyle y\,\mathrm{cosec}^{3}x=-2\,\mathrm{cosec}\,x+C\).Apply \(\displaystyle y=2\) when \(\displaystyle x=\frac{\pi}{2}\), where \(\displaystyle \sin\frac{\pi}{2}=1\) so \(\displaystyle \mathrm{cosec}\,\frac{\pi}{2}=1\): \[2=-2+C\qquad\Rightarrow\qquad C=4\]Hence \(\displaystyle y\,\mathrm{cosec}^{3}x=-2\,\mathrm{cosec}\,x+4\); multiplying by \(\displaystyle \sin^{3}x\): \[\boxed{\,y=4\sin^{3}x-2\sin^{2}x\,}\]

The slope of the tangent at \(\displaystyle (x,y)\) is \(\displaystyle \frac{dy}{dx}\), and it equals the sum of the coordinates: \[\frac{dy}{dx}=x+y\qquad\Rightarrow\qquad \frac{dy}{dx}-y=x\]This is linear with \(\displaystyle P=-1\), \(\displaystyle Q=x\): \[\mathrm{I.F.}=e^{\int(-1)dx}=e^{-x}\]Multiplying through: \[\frac{d}{dx}\left(y\,e^{-x}\right)=x\,e^{-x}\]Integrating by parts, \[\int x\,e^{-x}dx=-x\,e^{-x}+\int e^{-x}dx=-x\,e^{-x}-e^{-x}+C=-(x+1)e^{-x}+C\]So \(\displaystyle y\,e^{-x}=-(x+1)e^{-x}+C\), i.e. \(\displaystyle y=-(x+1)+C\,e^{x}\).The curve passes through the origin, so \(\displaystyle y=0\) when \(\displaystyle x=0\): \[0=-1+C\qquad\Rightarrow\qquad C=1\]The required curve is \[\boxed{\,x+y+1=e^{x}\,}\qquad\text{i.e.}\qquad y=e^{x}-x-1\]

Let the curve be \(\displaystyle y=f(x)\); the slope of the tangent at \(\displaystyle (x,y)\) is \(\displaystyle \frac{dy}{dx}\). "The sum of the coordinates exceeds the slope by $\displaystyle 5$" gives \[x+y=\frac{dy}{dx}+5\qquad\Rightarrow\qquad \frac{dy}{dx}-y=x-5\]This is linear with \(\displaystyle P=-1\), \(\displaystyle Q=x-5\): \[\mathrm{I.F.}=e^{\int(-1)dx}=e^{-x}\]Multiplying through: \[\frac{d}{dx}\left(y\,e^{-x}\right)=(x-5)e^{-x}\]Integrating by parts, \[\int(x-5)e^{-x}dx=-(x-5)e^{-x}+\int e^{-x}dx=-(x-5)e^{-x}-e^{-x}+C=-(x-4)e^{-x}+C\]So \(\displaystyle y\,e^{-x}=-(x-4)e^{-x}+C\), i.e. \(\displaystyle y=4-x+C\,e^{x}\).The curve passes through \(\displaystyle (0,2)\): \[2=4-0+C\qquad\Rightarrow\qquad C=-2\]The required curve is \[\boxed{\,y=4-x-2e^{x}\,}\]

An integrating factor is read off only from the standard form \(\displaystyle \frac{dy}{dx}+Py=Q\), so divide by \(\displaystyle x\) first (\(\displaystyle x\neq 0\)): \[x\frac{dy}{dx}-y=2x^{2}\qquad\Rightarrow\qquad \frac{dy}{dx}-\frac{1}{x}\,y=2x\]Here \(\displaystyle P=-\frac{1}{x}\), so \[\mathrm{I.F.}=e^{\int P\,dx}=e^{-\int\frac{1}{x}dx}=e^{-\log|x|}=e^{\log\frac{1}{|x|}}=\frac{1}{x}\]The correct option is \(\displaystyle \boxed{\text{(C) }\dfrac{1}{x}}\).

The equation is given with \(\displaystyle \frac{dx}{dy}\), so it is linear in \(\displaystyle x\) with \(\displaystyle y\) as the independent variable. Divide by \(\displaystyle 1-y^{2}\), which is non-zero since \(\displaystyle -1<y<1\): \[\left(1-y^{2}\right)\frac{dx}{dy}+yx=ay\qquad\Rightarrow\qquad \frac{dx}{dy}+\frac{y}{1-y^{2}}\,x=\frac{ay}{1-y^{2}}\]Here \(\displaystyle P_{1}=\frac{y}{1-y^{2}}\). Substituting \(\displaystyle t=1-y^{2}\), \(\displaystyle dt=-2y\,dy\): \[\int\frac{y}{1-y^{2}}dy=-\frac{1}{2}\int\frac{dt}{t}=-\frac{1}{2}\log t=-\frac{1}{2}\log\left(1-y^{2}\right)\]Hence \[\mathrm{I.F.}=e^{-\frac{1}{2}\log\left(1-y^{2}\right)}=\left(1-y^{2}\right)^{-1/2}=\frac{1}{\sqrt{1-y^{2}}}\]Since \(\displaystyle -1<y<1\) makes \(\displaystyle 1-y^{2}>0\), it is \(\displaystyle 1-y^{2}\) under the root and not \(\displaystyle y^{2}-1\).The correct option is \(\displaystyle \boxed{\text{(D) }\dfrac{1}{\sqrt{1-y^{2}}}}\).