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

Let $f: R \rightarrow R$ be a continuous function which satisfies $f(x) = \int_{0}^{x} f(t) \, dt$. Then,the value of $f(\log_{e} 5)$ is:

The solution of the differential equation $(y^{2}+2x) \frac{dy}{dx}=y$ satisfies $x=1, y=1$. Then the solution is

If the general solution of $(1+y^2) dx = (\operatorname{Tan}^{-1} y - x) dy$ is $x = f(y) + c e^{-\operatorname{Tan}^{-1} y}$,then $f(y) =$

$A$ function $y=f(x)$ satisfies $f(x) \sin 2x + \sin x - (1 + \cos^2 x) f'(x) = 0$ with the condition $f(0) = 0$. Then $f(\frac{\pi}{2})$ is equal to

The integrating factor of the differential equation $\frac{dy}{dx} = y \tan x - y^2 \sec x$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module