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. $(x y)^4+x^5 / 5=C$

Answer 1

Answer: (B)

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