Question

Solve: $\frac{dy}{dx}=\frac{x^2+xy}{x^2+y^2}$

• A. $\frac{c(x-y{)}^{2/3}}{(x^2+xy+y^2{)}^{1/6}}=exp\left[-\frac{1}{\sqrt{3}}ta{n}^{-1}\frac{x+2y}{x\sqrt{3}}\right]$ where exp $x=e^x$
• B. $c(x-y{)}^{4/3}(x^2+xy+y^2{)}^{1/4}=exp\left[-\frac{1}{\sqrt{3}}se{c}^{-1}\frac{x-2y}{x\sqrt{3}}\right]$ where exp $x=e^x$
• C. $\frac{c(x-y{)}^{2/3}}{(x^2+xy+y^2{)}^{1/6}}=exp\left[-\frac{1}{\sqrt{3}}se{c}^{-1}\frac{x-2y}{x\sqrt{5}}\right]$ where exp $x=e^x$
• D. none of these

Answer 1

Answer: (A)

$\frac{c(x-y{)}^{2/3}}{(x^2+xy+y^2{)}^{1/6}}=exp\left[-\frac{1}{\sqrt{3}}ta{n}^{-1}\frac{x+2y}{x\sqrt{3}}\right]$ where exp $x=e^x$

Given, $\frac{dy}{dx}=\frac{x^2+xy}{x^2+y^2}$
which is homogeneous differential eqn.
Put $y=vx$
$\frac{dy}{dx}=v+x\frac{dv}{dx}$
So, the given eqn becomes
$v+x\frac{dv}{dx}=\frac{1+v}{1+v^2}$
$\Rightarrow x\frac{dv}{dx}=\frac{1-v^3}{1+v^2}$
$\Rightarrow \frac{1+v^2}{1-v^3}dv=\frac{dx}{x}$
Integrating both sides,
$\int \frac{1+v^2}{(1-v)(1+v^2+v)}dv=\int \frac{dx}{x}$ .....(1)
Now, we will resolve $\frac{1+v^2}{(1-v)(1+v^2+v)}$ into partial fractions
$\frac{1+v^2}{(1-v)(1+v^2+v)}=\frac{A}{(1-v)}+\frac{Bv+C}{(1+v^2+v)}$ ....(2)
$\Rightarrow 1+v^2=A(1+v^2+v)+(Bv+C)(1-v)$
On comparing ,we get
$A-B=1;A+C=1;A+B-C=0$
Solving these eqns , we get
$A=\frac{2}{3},B=-\frac{1}{3},C=\frac{1}{3}$
Substituting these values in (2), we get
$\frac{1+v^2}{(1-v)(1+v^2+v)}=\frac{2}{3(1-v)}-\frac{1}{3}\frac{(v-1)}{(1+v^2+v)}$
So, eqn (1) becomes
$\int \frac{2}{3(1-v)}dv-\int \frac{1}{3}\frac{(v-1)}{(1+v^2+v)}dv=\int \frac{dx}{x}$
$\Rightarrow \frac{2}{3}\log |1-v|-\frac{1}{6}\int \frac{(2v+1-3)}{(1+v^2+v)}dv=\log x+\log C$
$\Rightarrow \frac{2}{3}\log |1-v|-\frac{1}{6}\int \frac{2v+1}{(1+v^2+v)}dv+\frac{1}{2}\int \frac{1}{(v+\frac{1}{2}{)}^2+(\frac{\sqrt{3}}{2}{)}^2}dv=\log x+\log C$
$\Rightarrow \frac{2}{3}\log |1-\frac{y}{x}|-\frac{1}{6}\log |v^2+v+1|+\frac{1}{\sqrt{3}}{\tan }^{-1}\left(\frac{2v+1}{\sqrt{3}}\right)=\log x+\log C$
$\Rightarrow \frac{2}{3}\log |x-y|-\frac{1}{6}\log |y^2+xy+x^2|+\frac{2}{3}\log x+\frac{1}{\sqrt{3}}{\tan }^{-1}\left(\frac{2y+x}{x\sqrt{3}}\right)=\frac{5}{3}\log x+\log C$
$\Rightarrow \log \frac{c(x-y{)}^{2/3}}{x(y^2+xy+x^2{)}^{1/6}}=-\frac{1}{\sqrt{3}}{\tan }^{-1}\left(\frac{2y+x}{x\sqrt{3}}\right)$
$\Rightarrow \frac{c(x-y{)}^{2/3}}{x(y^2+xy+x^2{)}^{1/6}}=exp[-\frac{1}{\sqrt{3}}{\tan }^{-1}\left(\frac{2y+x}{x\sqrt{3}}\right)]$