NCERT Solutions · Class 12 Mathematics Differential Equations

Write the equation as \(\displaystyle \frac{dy}{dx}=\frac{y-x}{x+y}\); numerator and denominator are homogeneous of degree $\displaystyle 1$, so the equation is homogeneous.Put \(\displaystyle y=vx\), \(\displaystyle \frac{dy}{dx}=v+x\frac{dv}{dx}\): \[v+x\frac{dv}{dx}=\frac{vx-x}{x+vx}=\frac{v-1}{1+v},\] \[x\frac{dv}{dx}=\frac{v-1-v\left(1+v\right)}{1+v}=-\frac{1+v^{2}}{1+v}.\] Separating the variables and splitting the left side, \[\frac{1+v}{1+v^{2}}\,dv=-\frac{dx}{x}\quad\Longrightarrow\quad\int\frac{dv}{1+v^{2}}+\int\frac{v\,dv}{1+v^{2}}=-\int\frac{dx}{x},\] \[\tan^{-1}v+\frac12\log\left(1+v^{2}\right)=-\log|x|+\mathrm C.\] 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: \[\tan^{-1}\frac yx+\frac12\log\left(x^{2}+y^{2}\right)=\mathrm C.\] Apply the condition \(\displaystyle y=1\) when \(\displaystyle x=1\): \[\tan^{-1}1+\frac12\log 2=\mathrm C\quad\Longrightarrow\quad\mathrm C=\frac{\pi}{4}+\frac12\log 2.\] The particular solution is \[\tan^{-1}\left(\frac yx\right)+\frac12\log\left(x^{2}+y^{2}\right)=\frac{\pi}{4}+\frac12\log 2,\] or equivalently \(\displaystyle \log\left(x^{2}+y^{2}\right)+2\tan^{-1}\left(\frac yx\right)=\frac{\pi}{2}+\log 2\).

Write the equation as \(\displaystyle \frac{dy}{dx}=-\frac{xy+y^{2}}{x^{2}}\); every term is homogeneous of degree $\displaystyle 2$, so the equation is homogeneous.Put \(\displaystyle y=vx\), \(\displaystyle \frac{dy}{dx}=v+x\frac{dv}{dx}\): \[v+x\frac{dv}{dx}=-\frac{vx^{2}+v^{2}x^{2}}{x^{2}}=-\left(v+v^{2}\right),\] \[x\frac{dv}{dx}=-2v-v^{2}=-v\left(v+2\right).\] Separating the variables, \[\frac{dv}{v\left(v+2\right)}=-\frac{dx}{x}.\] Resolve the left side into partial fractions, \(\displaystyle \frac{1}{v(v+2)}=\frac12\left(\frac1v-\frac{1}{v+2}\right)\). Integrating, \[\frac12\log\left|\frac{v}{v+2}\right|=-\log|x|+\mathrm C\quad\Longrightarrow\quad\log\left|\frac{v}{v+2}\right|=-2\log|x|+\mathrm{C}_{1}.\] Exponentiating, \(\displaystyle \frac{v}{v+2}=\frac{\mathrm A}{x^{2}}\). With \(\displaystyle v=\frac yx\), \[\frac{y/x}{\frac yx+2}=\frac{y}{y+2x}=\frac{\mathrm A}{x^{2}}\quad\Longrightarrow\quad x^{2}y=\mathrm A\left(y+2x\right).\] Apply \(\displaystyle y=1\) when \(\displaystyle x=1\): \(\displaystyle 1=\mathrm A(1+2)\), so \(\displaystyle \mathrm A=\frac13\). Then \(\displaystyle 3x^{2}y=y+2x\), i.e. \(\displaystyle y\left(3x^{2}-1\right)=2x\).The particular solution is \[y=\frac{2x}{3x^{2}-1},\qquad x\neq\pm\frac{1}{\sqrt3}.\]

Solve for the derivative: \(\displaystyle x\,dy=\left[y-x\sin^{2}\left(\frac yx\right)\right]dx\), so \[\frac{dy}{dx}=\frac yx-\sin^{2}\left(\frac yx\right),\] a function of \(\displaystyle y/x\) alone; the equation is homogeneous.Put \(\displaystyle y=vx\), \(\displaystyle \frac{dy}{dx}=v+x\frac{dv}{dx}\): \[v+x\frac{dv}{dx}=v-\sin^{2}v\quad\Longrightarrow\quad x\frac{dv}{dx}=-\sin^{2}v.\] Separating the variables, \[\mathrm{cosec}^{2}v\,dv=-\frac{dx}{x}.\] Since \(\displaystyle \int\mathrm{cosec}^{2}v\,dv=-\cot v\), integrating gives \[-\cot v=-\log|x|+\mathrm C\quad\Longrightarrow\quad\cot v=\log|x|+\mathrm{C}_{1}.\] Apply \(\displaystyle y=\frac{\pi}{4}\) when \(\displaystyle x=1\), so \(\displaystyle v=\frac{\pi}{4}\) and \(\displaystyle \cot\frac{\pi}{4}=1\): \[1=\log 1+\mathrm{C}_{1}=\mathrm{C}_{1}.\] The particular solution is \[\cot\left(\frac yx\right)=\log|x|+1=\log|ex|.\]

The equation is \(\displaystyle \frac{dy}{dx}=\frac yx-\mathrm{cosec}\left(\frac yx\right)\), a function of \(\displaystyle y/x\) alone, so it is homogeneous.Put \(\displaystyle y=vx\), \(\displaystyle \frac{dy}{dx}=v+x\frac{dv}{dx}\): \[v+x\frac{dv}{dx}=v-\mathrm{cosec}\,v\quad\Longrightarrow\quad x\frac{dv}{dx}=-\mathrm{cosec}\,v.\] Separating the variables, \[\sin v\,dv=-\frac{dx}{x}.\] Integrating, \[-\cos v=-\log|x|+\mathrm C\quad\Longrightarrow\quad\cos v=\log|x|+\mathrm{C}_{1}.\] Apply \(\displaystyle y=0\) when \(\displaystyle x=1\), so \(\displaystyle v=\frac01=0\) and \(\displaystyle \cos 0=1\): \[1=\log 1+\mathrm{C}_{1}=\mathrm{C}_{1}.\] The particular solution is \[\cos\left(\frac yx\right)=\log|x|+1=\log|ex|.\]

Solve for the derivative: \(\displaystyle 2x^{2}\frac{dy}{dx}=2xy+y^{2}\), so \[\frac{dy}{dx}=\frac{2xy+y^{2}}{2x^{2}},\] homogeneous of degree $\displaystyle 2$ in numerator and denominator.Put \(\displaystyle y=vx\), \(\displaystyle \frac{dy}{dx}=v+x\frac{dv}{dx}\): \[v+x\frac{dv}{dx}=\frac{2vx^{2}+v^{2}x^{2}}{2x^{2}}=v+\frac{v^{2}}{2},\] \[x\frac{dv}{dx}=\frac{v^{2}}{2}.\] Separating the variables, \[\frac{2\,dv}{v^{2}}=\frac{dx}{x}\quad\Longrightarrow\quad-\frac2v=\log|x|+\mathrm C.\] Substituting \(\displaystyle v=\frac yx\), \[-\frac{2x}{y}=\log|x|+\mathrm C.\] Apply \(\displaystyle y=2\) when \(\displaystyle x=1\): \(\displaystyle -\frac{2}{2}=\log 1+\mathrm C\), so \(\displaystyle \mathrm C=-1\). Then \[-\frac{2x}{y}=\log|x|-1\quad\Longrightarrow\quad\frac{2x}{y}=1-\log|x|.\] The particular solution is \[y=\frac{2x}{1-\log|x|},\qquad x\neq0,\;x\neq e.\]

In \(\displaystyle \frac{dx}{dy}=h\left(\frac xy\right)\) the dependent variable is \(\displaystyle x\) and the independent variable is \(\displaystyle y\), and the right side is a function of the ratio \(\displaystyle \frac xy\). The substitution must turn that ratio into a single new variable, so we set the ratio equal to \(\displaystyle v\): \[\frac xy=v\quad\Longrightarrow\quad x=vy,\qquad\frac{dx}{dy}=v+y\frac{dv}{dy}.\] The equation then becomes \(\displaystyle v+y\frac{dv}{dy}=h(v)\), i.e. \(\displaystyle \frac{dv}{h(v)-v}=\frac{dy}{y}\), which is variables-separable. (The substitution \(\displaystyle y=vx\) would leave the equation still containing both \(\displaystyle x\) and \(\displaystyle v\) unseparated, so it is the wrong one here.)Hence the correct option is (C) \(\displaystyle x=vy\).