The general solution of the differential equa | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(C) The given differential equation is $\frac{dy}{dx} + \frac{1}{x}y = x^2$. This is a linear differential equation of the form $\frac{dy}{dx} + Py =

Vedclass · Accepted Answer

$xy = \frac{x^4}{4} + C$

Vedclass · Answer

(C) The given differential equation is $\frac{dy}{dx} + \frac{1}{x}y = x^2$. This is a linear differential equation of the form $\frac{dy}{dx} + Py = Q$, where $P = \frac{1}{x}$ and $Q = x^2$. First, we find the integrating factor $(IF)$: $IF = e^{\int P dx} = e^{\int \frac{1}{x} dx} = e^{\ln x} = x$. The general solution is given by $y \cdot IF = \int (Q \cdot IF) dx + C$. Substituting the values, we get $y \cdot x = \int (x^2 \cdot x) dx + C$. $xy = \int x^3 dx + C$. Integrating $x^3$, we get $xy = \frac{x^4}{4} + C$.

The general solution of the differential equation $\frac{dy}{dx} + \frac{y}{x} = x^2$ is

Similar Questions

Solve the following differential equation: $\left(x^2+1\right) \frac{dy}{dx} + 4xy = \frac{1}{x^2+1}$

The integrating factor of the differential equation $x \frac{dy}{dx} + y \log x = x e^x \cdot x^{-1/2} \log x$ for $x > 0$ is:

Let $f(x)=x-1$ and $g(x)=e^x$ for $x \in R$. If $\frac{d y}{d x}=\left(e^{-2 \sqrt{x}} g(f(f(x)))-\frac{y}{\sqrt{x}}\right)$ and $y(0)=0$,then $y(1)$ is:

Let $y = y(x)$ be the solution curve of the differential equation $x(x^2 + e^x) dy + (e^x(x-2)y - x^3) dx = 0, x > 0$,passing through the point $(1, 0)$. Then $y(2)$ is equal to:

Solve the differential equation $\frac{dy}{dx} = \frac{1+y^2}{(\tan^{-1} y) - x}$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module