Find the general solution of the differential | vedclass

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

Question

(A) The given differential equation is a linear differential equation of the form $\frac{dy}{dx} + Py = Q$,where $P = -\frac{1}{x}$ and $Q = x^2$.Firs

Vedclass · Accepted Answer

$y = \frac{x^3}{3} + Cx$

Vedclass · Answer

(A) The given differential equation 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 $(I.F.)$:$I.F. = e^{\int P dx} = e^{\int -\frac{1}{x} dx} = e^{-\log x} = e^{\log(x^{-1})} = \frac{1}{x}$.The general solution is given by $y(I.F.) = \int (Q 	imes I.F.) dx + C$.Substituting the values:$y \cdot \frac{1}{x} = \int (x^2 \cdot \frac{1}{x}) dx + C$$\frac{y}{x} = \int x dx + C$$\frac{y}{x} = \frac{x^2}{2} + C$$y = \frac{x^3}{2} + Cx$.

Find the general solution of the differential equation: $\frac{dy}{dx} - \frac{y}{x} = x^2$

Similar Questions

The solution of the differential equation $(x + 2y^3) \frac{dy}{dx} - y = 0$ is

The integrating factor of the differential equation $(x^2 - 1)\frac{dy}{dx} + 2xy = x$ is

Let $y(x)$ be the solution of the differential equation $(x \log x) \frac{dy}{dx} + y = 2x \log x$,$(x \ge 1)$. Then $y(e)$ is equal to: $[y(1) = 0]$

Let the solution $y=y(x)$ of the differential equation $\frac{dy}{dx}-y=1+4 \sin x$ satisfy $y(\pi)=1$. Then $y\left(\frac{\pi}{2}\right)+10$ is equal to:

Observe the following statements:
$A$. Integrating factor of $\frac{dy}{dx} + y = x^2$ is $e^x$.
$R$. Integrating factor of $\frac{dy}{dx} + P(x)y = Q(x)$ is $e^{\int P(x) dx}$.
Then, the true statement among the following is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module