Question

$\frac{dy}{dx}=\frac{x^2+xy}{x^2+y^2}$
Solving this gives $c(x-y{)}^{2/3}(x^3+xy+y^3{)}^{1/5}=\left[\frac{1}{\sqrt{k}}ta{n}^{-1}\frac{x+2y}{x\sqrt{k}}\right]$ where $exp\left(x\right)=e^x$, then what is k?

• A. $2$
• B. $1$
• C. $3$
• D. $4$

Answer 1

Answer: (C)

$3$

Given differential equation is
$\frac{dy}{dx}=\frac{x^2+xy}{x^2+y^2}$
Put $y=vx$
$\frac{dy}{dx}=v+x\frac{dv}{dx}$
So, the given differential equation becomes
$v+x\frac{dv}{dx}=\frac{1+v}{1+v^2}$
$x\frac{dv}{dx}=\frac{1-v^3}{1+v^2}$
$\frac{(1+v^2)dv}{1-v^3}=\frac{dx}{x}$
$\frac{(1+v^2)dv}{(1-v)(1+v^2+v)}=\frac{dx}{x}$
Integrating both sides ,
$\int \frac{(1+v^2)dv}{(1-v)(1+v^2+v)}=\int \frac{dx}{x}$ .....(1)
Now, we will resolve $\frac{(1+v^2)}{(1-v)(1+v^2+v)}$ into partial fractions
$\frac{(1+v^2)}{(1-v)(1+v^2+v)}=\frac{A}{(1-v)}+\frac{Bv+C}{(1+v^2+v)}$
$\Rightarrow 1+v^2=A(1+v^2+v)+(Bv+C)(1-v)$
On comparing , we get
$A-B=1$
$A+C=1$
$A+B-C=0$
Solving these equations, we get
$A=\frac{2}{3},B=-\frac{1}{3},C=\frac{1}{3}$
$\int \frac{(1+v^2)dv}{(1-v)(1+v^2+v)}=\int \frac{2}{3(1-v)}dv-\int \frac{v-1}{3(1+v^2+v)}dv$
$\int \frac{(1+v^2)dv}{(1-v)(1+v^2+v)}=\frac{2}{3}\log (1-v)-\frac{1}{6}\int \frac{2v+1-3}{1+v^2+v}dv$
$\int \frac{(1+v^2)dv}{(1-v)(1+v^2+v)}=\frac{2}{3}\log (1-v)-\frac{1}{6}\int \frac{2v+1}{1+v^2+v}dv+\frac{1}{2}\int \frac{1}{1+v^2+v}dv$
Put $v^2+v+1=t$
$\Rightarrow (2v+1)dv=dt$
$\int \frac{(1+v^2)dv}{(1-v)(1+v^2+v)}=\frac{2}{3}\log (1-v)-\frac{1}{6}\int \frac{1}{t}dt+\frac{1}{2}\int \frac{1}{(v+\frac{1}{2}{)}^2+{\left(\frac{\sqrt{3}}{2}\right)}^2}dv$
$\Rightarrow \int \frac{(1+v^2)dv}{(1-v)(1+v^2+v)}=\frac{2}{3}\log (1-v)-\frac{1}{6}\log t+\frac{1}{\sqrt{3}}{\tan }^{-1}\left(\frac{2v+1}{\sqrt{3}}\right)$
$\int \frac{(1+v^2)dv}{(1-v)(1+v^2+v)}=\frac{2}{3}\log (x-y)-\frac{2}{3}\log x-\frac{1}{6}\log (x^2+y^2+xy)+\frac{1}{\sqrt{3}}{\tan }^{-1}\left(\frac{2y+x}{x\sqrt{3}}\right)$
Put this value in (1), we get
$\frac{2}{3}\log (x-y)-\frac{2}{3}\log x-\frac{1}{6}\log (x^2+y^2+xy)+\frac{1}{\sqrt{3}}{\tan }^{-1}\left(\frac{2y+x}{x\sqrt{3}}\right)=\log x+\log c$
$\Rightarrow \frac{1}{\sqrt{3}}{\tan }^{-1}\left(\frac{2y+x}{x\sqrt{3}}\right)=\frac{5}{3}\log x-\frac{2}{3}\log (x-y)+\frac{1}{6}\log (x^2+y^2+xy)+\log c$
$\Rightarrow \log \frac{c{x}^{5/3}(x^2+y^2+xy{)}^{1/6}}{(x-y{)}^{2/3}}=\frac{1}{\sqrt{3}}{\tan }^{-1}\left(\frac{2y+x}{x\sqrt{3}}\right)$
On comparing, we get $k=3$