Question

A differential equation of the form $\frac{dy}{dx}+Py=Q$ ...(*)

where P & Q are functions of x. The number $e^{\int Pdx}$ when multiplied to R.H.S of (*) make it differential coefficient of a function of x & y, is called the integrating factor of the differential equation given by (*). Further the equation $\frac{dy}{dx}+Py=Q{y}^n$ where P, Q are functions of x is reducible to linear form by substituting $y^{-n+1}$ as new dependent variable.

On the basis of the above information answer the following question.

The integrating factor of differential equation $\frac{dy}{dx}+\frac{y}{x}=y^2$ is

• A. $\frac{1}{y}$
• B. $-\frac{1}{y}$
• C. $\frac{1}{x}$
• D. $-\frac{1}{x}$

Answer 1

The correct option is C $\frac{1}{x}$

$\frac{dy}{dx}+\frac{y}{x}=y^2$ (*)

$\Rightarrow$ $\frac{dy}{dx}\frac{1}{y^2}+\frac{1}{xy}=1$

$\Rightarrow$ $\frac{1}{y^2}\frac{dy}{dx}+\frac{1}{y}\frac{1}{x}=1$

Substitute $\frac{1}{y}=t$

$\Rightarrow$ $-\frac{1}{y^2}\frac{dy}{dx}=\frac{dt}{dx}$

$\Rightarrow$ $\frac{dy}{dx}=-y^2\frac{dt}{dx}=\frac{-1}{t^2}\frac{dt}{dx}$

$\therefore$ Equation (*) reduces to,

$\Rightarrow$ $-\frac{dt}{dx}+\frac{1}{x}t=1$

$\Rightarrow$ $\frac{dt}{dx}-\frac{1}{x}t=-1$ which is linear in t

$\therefore$ I.F. $=e^{\int \frac{-1}{x}dx}=e^{-\log x}=\frac{1}{x}$