Question

The equation of curve satisfying differential equation $\frac{dy}{dx}=x\ln x$ and passing through the point $(e,1)$ is

• A. $y=\frac{x^2}{2}\ln x-\frac{x^2}{4}+1$
• B. $y=x^2\ln x-\frac{x^2}{2}+1-\frac{e^2}{4}$
• C. $y=\frac{x^2}{2}\ln x-\frac{x^2}{4}+\frac{e^2}{4}$
• D. $y=\frac{x^2}{2}\ln x-\frac{x^2}{4}+1-\frac{e^2}{4}$

Answer 1

The correct option is D $y=\frac{x^2}{2}\ln x-\frac{x^2}{4}+1-\frac{e^2}{4}$

$dy=x\ln xdx$
Integrating both sides, we get
$y=\frac{x^2}{2}\ln x-\int \frac{x}{2}dx$
$y=\frac{x^2}{2}\ln x-\frac{x^2}{4}+c$
Curve passes through point $(e,1)$
$\Rightarrow 1=\frac{e^2}{2}\ln e-\frac{e^2}{4}+c$
$\Rightarrow c=1-\frac{e^2}{4}$
$\therefore$ Equation of curve is
$y=\frac{x^2}{2}\ln x-\frac{x^2}{4}+1-\frac{e^2}{4}$