Question

Equation of curve which passes through point $(1,1)$ and satisfies the differential equation $3x{y}^{2}dy=(x^2+2y^3)dx$ is

• A. $y^3=x\ln \left(e{x}^2\right)$
• B. $y=\ln \left(xe\right)$
• C. $y^6=x^4\ln \left(x^{2}e\right)$
• D. $y^3=x^2\ln \left(ex\right)$

Answer 1

The correct option is D $y^3=x^2\ln \left(ex\right)$

$3x{y}^{2}dy=(x^2+2y^3)dx$
$\Rightarrow 3y^2\frac{dy}{dx}-\frac{2}{x}{y}^3=x\dots \left(1\right)$
Put $y^3=t$
$\Rightarrow 3y^2\frac{dy}{dx}=\frac{dt}{dx}$
From equation $\left(1\right)$
$\Rightarrow \frac{dt}{dx}-\frac{2t}{x}=x$
$I.F.=e^{\int \frac{-2}{x}dx}=e^{\ln \left(\frac{1}{x^2}\right)}=\frac{1}{x^2}$
$y^3\cdot \frac{1}{x^2}=\int x\cdot \frac{1}{x^2}dx$
$\Rightarrow \frac{y^3}{x^2}=\ln x+c$
Equation satisfies point $(1,1)$
$\Rightarrow 1=c$
$\Rightarrow \frac{y^3}{x^2}=\ln x+1$
$\Rightarrow y^3=x^2\ln \left(ex\right)$