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 of the form $\frac{dy}{dx} + Py = Q$,where $P = 3$ and $Q = e^{-2x}$.First,we find the integrating factor $(I.F

Vedclass · Accepted Answer

$y = e^{-2x} + Ce^{-3x}$

Vedclass · Answer

(A) The given differential equation is of the form $\frac{dy}{dx} + Py = Q$,where $P = 3$ and $Q = e^{-2x}$.First,we find the integrating factor $(I.F.)$:$I.F. = e^{\int P dx} = e^{\int 3 dx} = e^{3x}$.The general solution is given by:$y(I.F.) = \int (Q 	imes I.F.) dx + C$.Substituting the values:$y e^{3x} = \int (e^{-2x} 	imes e^{3x}) dx + C$$y e^{3x} = \int e^{x} dx + C$$y e^{3x} = e^{x} + C$.Dividing both sides by $e^{3x}$:$y = \frac{e^{x}}{e^{3x}} + \frac{C}{e^{3x}}$$y = e^{-2x} + Ce^{-3x}$.

Find the general solution of the differential equation: $\frac{dy}{dx} + 3y = e^{-2x}$

Similar Questions

Let $f: R \rightarrow R$ be a continuous function satisfying $f(x)=x+\int_0^x f(t) dt$,for all $x \in R$. Then,the number of elements in the set $S=\{x \in R: f(x)=0\}$ is

The general solution of the differential equation $(y^2+x+1) dy = (y+1) dx$ is

The solution of the differential equation $(1+y^2)+(x-e^{\tan ^{-1} y}) \frac{dy}{dx}=0$ is

The integrating factor of the differential equation $(1+y+x^{2}y)dx + (x+x^{3})dy = 0$ is

Which of the following equation$(s)$ is/are linear?

Vedclass Test Series

Exam Paper Generator

Online Exam Module