Question

The solution of differential equation $\frac{xdy}{dx}+y=xlogx$ is

• A. $xy=\frac{x^2}{2}logx-\frac{x^2}{4}$
• B. $xy=\frac{x^2}{2}logx-\frac{x^2}{4}$
• C. $xy=\frac{x^2}{2}logx-\frac{x^2}{4}+c,whereisaconstant$
• D. $Noneoftheabove$

Answer 1

The correct option is C $xy=\frac{x^2}{2}logx-\frac{x^2}{4}+c,whereisaconstant$

The given equation is $\frac{xdy}{dx}+y=xlogx$
Now you can see there is no multiplication of y and $\frac{dy}{dx}$, neither there is any transcendental function of y. Also dependent variable and y occur to its first degree. So this may be linear differential equation of first order.

If I write the given equation as $\frac{dy}{dx}+\frac{y}{x}=logx$
Then it matches to $\frac{dy}{dx}+P\left(x\right)y=Q\left(x\right)$
Where $P\left(x\right)=\frac{1}{x},Q\left(x\right)=log\left(x\right)$
So this one is a linear differential equation of first order. Whose solution is.
y X (I.F) = $\int$ Q(I.F.)dx + C
Where I.F. = $e^{\int Pdx}$
So I.F. = $e^{\int \frac{1}{x}dx}=e^{logx}=x$
So solution is xy = $\int$ x log x dx + C
$\Rightarrow xy=\frac{x^2}{2}logx-\frac{x^2}{4}+C$