Question

What is the solution of the differential equation $x\frac{dy}{dx}+y=x\ln x$ ?

• A. $xy=\frac{x^2}{2}\ln x-\frac{x^2}{4}$
• B. $xy=\frac{x^2}{2}\ln x-\frac{x^2}{4}+c$, where c is an arbitrary constant
• C. $xy=\frac{x^2}{2}\ln x-\frac{x^3}{4}+c$, where c is an arbitrary constant
• D. $xy=\frac{x^3}{6}\ln x-\frac{x^2}{4}+c$, where c is an arbitrary constant

Answer 1

The correct option is B $xy=\frac{x^2}{2}\ln x-\frac{x^2}{4}+c$, where c is an arbitrary constant

The given DE can be written as $\frac{dy}{dx}+\frac{y}{x}=\ln x$
Then it matches to $\frac{dy}{dx}+P\left(x\right)y=Q\left(x\right)$
Where $P\left(x\right)=\frac{1}{x},Q\left(x\right)=\ln \left(x\right)$
So this one is a linear differential equation of first order. Whose solution is.
$y\cdot \text{(I.F)}=\int \text{Q(I.F.)}dx+C$
Where $\text{I.F.}=e^{\int Pdx}$
So I.F. = $e^{\int \frac{1}{x}dx}=e^{\ln x}=x$
So solution is $xy=\int x\ln xdx+C$
$\Rightarrow xy=\frac{x^2}{2}\ln x-\frac{x^2}{4}+C$, where C is an arbitrary constant