Search any question & find its solution
Question:
Answered & Verified by Expert
The general solution of the differential equation $\frac{d y}{d x}+\frac{y}{x}=x^2$ is
Options:
Solution:
1083 Upvotes
Verified Answer
The correct answer is:
$x y=\frac{x^4}{4}+c$
$\frac{d y}{d x}+\frac{y}{x}=x^2$
$\begin{aligned} & \Rightarrow \quad \frac{d y}{d x}+\frac{1}{x} y=x^2 \\ & \Rightarrow \quad P=\frac{1}{x}, Q=x^2 \\ & \Rightarrow \quad I F=e^{\int P d x}=e^{\int \frac{1}{x} d x}=e^{\ln x}=x \\ & \Rightarrow \quad y \cdot I F=\int Q \cdot I F d x+C \\ & \Rightarrow \quad x y=\int x^2 \cdot x d x+C \\ & \therefore \quad x y=\frac{x^4}{4}+C .\end{aligned}$
$\begin{aligned} & \Rightarrow \quad \frac{d y}{d x}+\frac{1}{x} y=x^2 \\ & \Rightarrow \quad P=\frac{1}{x}, Q=x^2 \\ & \Rightarrow \quad I F=e^{\int P d x}=e^{\int \frac{1}{x} d x}=e^{\ln x}=x \\ & \Rightarrow \quad y \cdot I F=\int Q \cdot I F d x+C \\ & \Rightarrow \quad x y=\int x^2 \cdot x d x+C \\ & \therefore \quad x y=\frac{x^4}{4}+C .\end{aligned}$
Looking for more such questions to practice?
Download the MARKS App - The ultimate prep app for IIT JEE & NEET with chapter-wise PYQs, revision notes, formula sheets, custom tests & much more.