Question

The solution of $y^{5}x+y-x\frac{dy}{dx}=0$ is -

• A. $x^4/4+1/5(x/y{)}^5=C$
• B. $x^5/5+(1/4)(x/y{)}^4=C$
• C. $(x/y{)}^5+x^4/4=C$
• D. $(xy{)}^4+x^5/5=C$

Answer 1

Answer: (B)

$y^{5}x+y-x\frac{dy}{dx}=0$
$\Rightarrow \frac{dy}{dx}-\frac{y}{x}=y^5$
$\Rightarrow \frac{1}{y^5}\frac{dy}{dx}-\frac{1}{x}\frac{1}{y^4}=1$
Let $\frac{1}{y^4}=t\Rightarrow \frac{-4}{y^5}\frac{dy}{dx}=\frac{dt}{dx}$
$\therefore -\frac{1}{4}\frac{dt}{dx}-\frac{t}{x}=1$
$\Rightarrow \frac{dt}{dx}+\frac{4t}{x}=-4$
I.F. $=e^{\int \frac{4}{x}dx}=e^{4\ln x}=x^4$
so solution is t.x $=-\int 4\cdot x^{4}dx+c$
$\Rightarrow \frac{1}{4}{\left(\frac{x}{y}\right)}^4+\frac{x^5}{5}=c$