Question

The solution of the differential equation $ydx+(x-y^3)dy=0$ is

• A. $xy=\frac{1}{3}{y}^3+C$
• B. $xy=y^4+C$
• C. $y^4=4xy+C$
• D. $4y=y^3+C$

Answer 1

Answer: (C)

Given $ydx+(x-y^3)dy=0$
$\Rightarrow y\frac{dx}{dy}+x-y^3=0\Rightarrow \frac{dx}{dy}+\frac{1}{y}x=y^2$
compare this equation to general equation
i.e. $\frac{dx}{dy}+Px=Q\Rightarrow P=\frac{1}{y},Q=y^2$
$\therefore$ I.f .=$e^{\int Pdy}=e^{\int \frac{1}{y}dy}=e^{logy}=y$
x × I.f. =$\int Q\times I.f.dy+C_1$
$\Rightarrow x.y=\int y^2.ydy+C_1\Rightarrow xy=\frac{y^4}{4}+C_1$
$\Rightarrow 4xy=y^4+4C_1$
$\Rightarrow y^4=4xy-4C_1\Rightarrow y^4=4xy+C,$
where C =-4C${}_1$