NCERT Solutions · Class 12 Mathematics Differential Equations

Write the equation as \(\displaystyle \frac{dy}{dx}=\frac{x^{2}+y^{2}}{x^{2}+xy}\). Numerator and denominator are both homogeneous of degree $\displaystyle 2$, so the right side is a function of \(\displaystyle y/x\) alone and the equation is homogeneous.Substitute \(\displaystyle y=vx\), so that \(\displaystyle \frac{dy}{dx}=v+x\frac{dv}{dx}\): \[v+x\frac{dv}{dx}=\frac{x^{2}+v^{2}x^{2}}{x^{2}+vx^{2}}=\frac{1+v^{2}}{1+v}.\] Therefore \[x\frac{dv}{dx}=\frac{1+v^{2}}{1+v}-v=\frac{1+v^{2}-v-v^{2}}{1+v}=\frac{1-v}{1+v}.\] Separating the variables, \[\frac{1+v}{1-v}\,dv=\frac{dx}{x}.\] The left integrand is improper in \(\displaystyle v\); rewrite \(\displaystyle 1+v=-(1-v)+2\), so that \(\displaystyle \frac{1+v}{1-v}=-1+\frac{2}{1-v}\). Integrating, \[-v-2\log|1-v|=\log|x|+\mathrm C.\] Put \(\displaystyle v=\frac{y}{x}\) and use \(\displaystyle \log\left|1-\frac yx\right|=\log|x-y|-\log|x|\): \[-\frac yx-2\log|x-y|+2\log|x|=\log|x|+\mathrm C,\] \[\log\frac{(x-y)^{2}}{|x|}=-\frac yx-\mathrm C.\] Exponentiating, the general solution is \[(x-y)^{2}e^{y/x}=\mathrm{C}_{1}\,x,\qquad\text{equivalently}\qquad (x-y)^{2}=\mathrm{C}_{1}\,x\,e^{-y/x},\] with \(\displaystyle \mathrm{C}_{1}\) an arbitrary constant and \(\displaystyle x\neq0\).

Here \(\displaystyle \frac{dy}{dx}=\frac{x+y}{x}=1+\frac yx\), a function of \(\displaystyle y/x\) alone, so the equation is homogeneous.Substitute \(\displaystyle y=vx\), so \(\displaystyle \frac{dy}{dx}=v+x\frac{dv}{dx}\): \[v+x\frac{dv}{dx}=1+v\quad\Longrightarrow\quad x\frac{dv}{dx}=1.\] Separating the variables, \(\displaystyle dv=\frac{dx}{x}\), and integrating, \[v=\log|x|+\mathrm C.\] Replacing \(\displaystyle v=\frac yx\), the general solution is \[y=x\left(\log|x|+\mathrm C\right),\qquad x\neq0.\]

Rearranging, \(\displaystyle (x-y)\,dy=(x+y)\,dx\), i.e. \(\displaystyle \frac{dy}{dx}=\frac{x+y}{x-y}\). Numerator and denominator are homogeneous of degree $\displaystyle 1$, so the equation is homogeneous.Put \(\displaystyle y=vx\), \(\displaystyle \frac{dy}{dx}=v+x\frac{dv}{dx}\): \[v+x\frac{dv}{dx}=\frac{x+vx}{x-vx}=\frac{1+v}{1-v},\] \[x\frac{dv}{dx}=\frac{1+v-v(1-v)}{1-v}=\frac{1+v^{2}}{1-v}.\] Separating the variables and splitting the left side into the two standard integrals, \[\frac{1-v}{1+v^{2}}\,dv=\frac{dx}{x}\quad\Longrightarrow\quad\int\frac{dv}{1+v^{2}}-\int\frac{v\,dv}{1+v^{2}}=\int\frac{dx}{x}.\] The second integral needs \(\displaystyle t=1+v^{2}\), giving \(\displaystyle \frac12\log(1+v^{2})\). Hence \[\tan^{-1}v-\frac12\log\left(1+v^{2}\right)=\log|x|+\mathrm C.\] With \(\displaystyle v=\frac yx\), \(\displaystyle \frac12\log\left(1+\frac{y^{2}}{x^{2}}\right)=\frac12\log\left(x^{2}+y^{2}\right)-\log|x|\), and the \(\displaystyle \log|x|\) terms cancel: \[\tan^{-1}\frac yx-\frac12\log\left(x^{2}+y^{2}\right)=\mathrm C.\] So the general solution is \[\tan^{-1}\left(\frac yx\right)=\frac12\log\left(x^{2}+y^{2}\right)+\mathrm C.\]

Write the equation as \(\displaystyle \frac{dy}{dx}=\frac{y^{2}-x^{2}}{2xy}\). Every term is homogeneous of degree $\displaystyle 2$, so the equation is homogeneous.Put \(\displaystyle y=vx\), \(\displaystyle \frac{dy}{dx}=v+x\frac{dv}{dx}\): \[v+x\frac{dv}{dx}=\frac{v^{2}x^{2}-x^{2}}{2vx^{2}}=\frac{v^{2}-1}{2v},\] \[x\frac{dv}{dx}=\frac{v^{2}-1-2v^{2}}{2v}=-\frac{1+v^{2}}{2v}.\] Separating the variables, \[\frac{2v}{1+v^{2}}\,dv=-\frac{dx}{x}.\] The left side is exactly \(\displaystyle \frac{d}{dv}\log\left(1+v^{2}\right)\), so integrating gives \[\log\left(1+v^{2}\right)=-\log|x|+\mathrm C.\] Substituting \(\displaystyle v=\frac yx\), \[\log\frac{x^{2}+y^{2}}{x^{2}}+\log|x|=\mathrm C\quad\Longrightarrow\quad\log\frac{x^{2}+y^{2}}{|x|}=\mathrm C.\] Exponentiating, the general solution is \[x^{2}+y^{2}=\mathrm{C}_{1}x,\] a family of circles through the origin with centres on the \(\displaystyle x\)-axis (\(\displaystyle x\neq0\)).

Divide by \(\displaystyle x^{2}\;(x\neq0)\): \(\displaystyle \frac{dy}{dx}=1-2\left(\frac yx\right)^{2}+\frac yx\), a function of \(\displaystyle y/x\) alone, so the equation is homogeneous.Put \(\displaystyle y=vx\), \(\displaystyle \frac{dy}{dx}=v+x\frac{dv}{dx}\): \[v+x\frac{dv}{dx}=1-2v^{2}+v\quad\Longrightarrow\quad x\frac{dv}{dx}=1-2v^{2}.\] Separating the variables, \[\frac{dv}{1-2v^{2}}=\frac{dx}{x}.\] For the left side, factor out the $\displaystyle 2$ so as to reach the standard form \(\displaystyle \int\frac{dt}{a^{2}-t^{2}}=\frac{1}{2a}\log\left|\frac{a+t}{a-t}\right|\): \[\int\frac{dv}{1-2v^{2}}=\frac12\int\frac{dv}{\left(\dfrac{1}{\sqrt2}\right)^{2}-v^{2}}=\frac12\cdot\frac{\sqrt2}{2}\log\left|\frac{\dfrac{1}{\sqrt2}+v}{\dfrac{1}{\sqrt2}-v}\right|=\frac{1}{2\sqrt2}\log\left|\frac{1+\sqrt2\,v}{1-\sqrt2\,v}\right|.\] Hence \[\frac{1}{2\sqrt2}\log\left|\frac{1+\sqrt2\,v}{1-\sqrt2\,v}\right|=\log|x|+\mathrm C.\] Putting \(\displaystyle v=\frac yx\) and multiplying numerator and denominator inside the logarithm by \(\displaystyle x\), the general solution is \[\frac{1}{2\sqrt2}\log\left|\frac{x+\sqrt2\,y}{x-\sqrt2\,y}\right|=\log|x|+\mathrm C.\]

Write the equation as \(\displaystyle \frac{dy}{dx}=\frac{y+\sqrt{x^{2}+y^{2}}}{x}\); both parts of the numerator are homogeneous of degree $\displaystyle 1$, so the equation is homogeneous.Put \(\displaystyle y=vx\), \(\displaystyle \frac{dy}{dx}=v+x\frac{dv}{dx}\). The step to watch is the surd: \(\displaystyle \sqrt{x^{2}+v^{2}x^{2}}=|x|\sqrt{1+v^{2}}\). Take \(\displaystyle x>0\) (for \(\displaystyle x<0\) only the sign of the arbitrary constant changes), so \[v+x\frac{dv}{dx}=\frac{vx+x\sqrt{1+v^{2}}}{x}=v+\sqrt{1+v^{2}},\] \[x\frac{dv}{dx}=\sqrt{1+v^{2}}.\] Separating the variables, \[\frac{dv}{\sqrt{1+v^{2}}}=\frac{dx}{x}.\] Using \(\displaystyle \int\frac{dv}{\sqrt{v^{2}+a^{2}}}=\log\left|v+\sqrt{v^{2}+a^{2}}\right|\), \[\log\left|v+\sqrt{1+v^{2}}\right|=\log|x|+\mathrm C.\] Exponentiating, \(\displaystyle v+\sqrt{1+v^{2}}=\mathrm{C}_{1}x\). Substituting \(\displaystyle v=\frac yx\) and \(\displaystyle \sqrt{1+v^{2}}=\frac{\sqrt{x^{2}+y^{2}}}{x}\), \[\frac{y}{x}+\frac{\sqrt{x^{2}+y^{2}}}{x}=\mathrm{C}_{1}x.\] The general solution is \[y+\sqrt{x^{2}+y^{2}}=\mathrm{C}_{1}x^{2}.\]

Write the equation as \[\frac{dy}{dx}=\frac{y\left\{x\cos\left(\frac yx\right)+y\sin\left(\frac yx\right)\right\}}{x\left\{y\sin\left(\frac yx\right)-x\cos\left(\frac yx\right)\right\}}.\] Numerator and denominator are homogeneous of degree $\displaystyle 2$, so the equation is homogeneous.Put \(\displaystyle y=vx\), \(\displaystyle \frac{dy}{dx}=v+x\frac{dv}{dx}\). The numerator becomes \(\displaystyle vx\left(x\cos v+vx\sin v\right)=vx^{2}\left(\cos v+v\sin v\right)\) and the denominator \(\displaystyle x\left(vx\sin v-x\cos v\right)=x^{2}\left(v\sin v-\cos v\right)\). Hence \[v+x\frac{dv}{dx}=\frac{v\left(\cos v+v\sin v\right)}{v\sin v-\cos v},\] \[x\frac{dv}{dx}=\frac{v\cos v+v^{2}\sin v-v\left(v\sin v-\cos v\right)}{v\sin v-\cos v}=\frac{2v\cos v}{v\sin v-\cos v}.\] Note how the \(\displaystyle v^{2}\sin v\) terms cancel, leaving only \(\displaystyle 2v\cos v\). Separating the variables, \[\frac{v\sin v-\cos v}{v\cos v}\,dv=\frac{2\,dx}{x}\quad\Longrightarrow\quad\left(\tan v-\frac1v\right)dv=\frac{2\,dx}{x}.\] Integrating, with \(\displaystyle \int\tan v\,dv=-\log|\cos v|\), \[-\log|\cos v|-\log|v|=2\log|x|+\mathrm C,\] \[\log\left|\frac{1}{v\cos v}\right|=\log x^{2}+\mathrm C\quad\Longrightarrow\quad v\,x^{2}\cos v=\mathrm{C}_{1}.\] Substituting \(\displaystyle v=\frac yx\), so that \(\displaystyle v x^{2}=xy\), the general solution is \[xy\cos\left(\frac yx\right)=\mathrm{C}.\]

Write the equation as \(\displaystyle \frac{dy}{dx}=\frac{y-x\sin\left(\frac yx\right)}{x}=\frac yx-\sin\left(\frac yx\right)\), a function of \(\displaystyle y/x\) alone, so the equation is homogeneous.Put \(\displaystyle y=vx\), \(\displaystyle \frac{dy}{dx}=v+x\frac{dv}{dx}\): \[v+x\frac{dv}{dx}=v-\sin v\quad\Longrightarrow\quad x\frac{dv}{dx}=-\sin v.\] Separating the variables, \[\mathrm{cosec}\,v\,dv=-\frac{dx}{x}.\] Using \(\displaystyle \int\mathrm{cosec}\,v\,dv=\log\left|\tan\frac v2\right|\), \[\log\left|\tan\frac v2\right|=-\log|x|+\mathrm C\quad\Longrightarrow\quad\log\left|x\tan\frac v2\right|=\mathrm C.\] Exponentiating and substituting \(\displaystyle v=\frac yx\), the general solution is \[x\tan\left(\frac{y}{2x}\right)=\mathrm{C}_{1},\qquad x\neq0.\]

Collect the \(\displaystyle dy\) terms: \(\displaystyle y\,dx=\left\{2x-x\log\left(\frac yx\right)\right\}dy\), so \[\frac{dy}{dx}=\frac{y}{x\left\{2-\log\left(\frac yx\right)\right\}},\] which depends only on \(\displaystyle y/x\); the equation is homogeneous.Put \(\displaystyle y=vx\), \(\displaystyle \frac{dy}{dx}=v+x\frac{dv}{dx}\): \[v+x\frac{dv}{dx}=\frac{vx}{x\left(2-\log v\right)}=\frac{v}{2-\log v},\] \[x\frac{dv}{dx}=\frac{v-v\left(2-\log v\right)}{2-\log v}=\frac{v\left(\log v-1\right)}{2-\log v}.\] Separating the variables, \[\frac{2-\log v}{v\left(\log v-1\right)}\,dv=\frac{dx}{x}.\] Substitute \(\displaystyle t=\log v-1\), so \(\displaystyle dt=\frac{dv}{v}\) and \(\displaystyle 2-\log v=1-t\). The left side becomes \[\int\frac{1-t}{t}\,dt=\int\left(\frac1t-1\right)dt=\log|t|-t.\] Hence \[\log\left|\log v-1\right|-\left(\log v-1\right)=\log|x|+\mathrm C,\] \[\log\left|\log v-1\right|=\log|x|+\log v+\mathrm{C}_{1}.\] With \(\displaystyle v=\frac yx\) the two logarithms on the right combine: \(\displaystyle \log|x|+\log\frac yx=\log|y|\). So \[\log\left|\log\frac yx-1\right|=\log|y|+\mathrm{C}_{1}.\] Exponentiating, the general solution is \[\log\left(\frac yx\right)-1=\mathrm{C}y,\qquad\text{i.e.}\qquad\log\left(\frac yx\right)=1+\mathrm{C}y.\]

Every occurrence of the variables is through \(\displaystyle x/y\), so this equation is homogeneous but in the \(\displaystyle x/y\) form; the right substitution is \(\displaystyle x=vy\), not \(\displaystyle y=vx\). Solving for \(\displaystyle \frac{dx}{dy}\), \[\frac{dx}{dy}=-\frac{e^{x/y}\left(1-\frac xy\right)}{1+e^{x/y}}.\] Put \(\displaystyle x=vy\), so \(\displaystyle \frac{dx}{dy}=v+y\frac{dv}{dy}\): \[v+y\frac{dv}{dy}=-\frac{e^{v}(1-v)}{1+e^{v}},\] \[y\frac{dv}{dy}=\frac{-e^{v}+ve^{v}-v\left(1+e^{v}\right)}{1+e^{v}}=\frac{-e^{v}+ve^{v}-v-ve^{v}}{1+e^{v}}=-\frac{v+e^{v}}{1+e^{v}}.\] Separating the variables, \[\frac{\left(1+e^{v}\right)dv}{v+e^{v}}=-\frac{dy}{y}.\] The numerator on the left is exactly the derivative of the denominator, so integrating gives \[\log\left|v+e^{v}\right|=-\log|y|+\mathrm C\quad\Longrightarrow\quad y\left(v+e^{v}\right)=\mathrm{C}_{1}.\] Substituting \(\displaystyle v=\frac xy\), \[y\left(\frac xy+e^{x/y}\right)=\mathrm{C}_{1}.\] The general solution is \[x+y\,e^{x/y}=\mathrm{C},\qquad y\neq0.\]