Question

We know that for a linear differential equation of first order, I.F = $e^{\int Pdx}$. Then value of p for $\frac{xdy}{dx}+x^{2}y=xlogx$ is

• A. $x^2$
• B. $x$
• C. $\frac{1}{x}$
• D. $Noneofthese$

Answer 1

The correct option is B $x$

Given equation is $\frac{xdy}{dx}+x^{2}y=xlogx$ Now the coefficient of y is P as per the standard form of linear differential equation. So p = $x^2$ Be careful guys. This is not the correct way of doing it.The standard form of Linear Differential equation is $\frac{dy}{dx}+P\left(x\right)y=Q\left(x\right)$ As per the standard form the coefficient of $\frac{dy}{dx}$ should be 1. Only then you can compare any equation to the standard form. Here the coefficient of $\frac{dy}{dx}$ is x and not 1. So first divide the entire equation by x.
It becomes $\frac{dy}{dx}+xy=logx$
Now this is comparable to standard form $\frac{dy}{dx}+P\left(x\right)y=Q\left(x\right)$
Here P = x
Q = log x