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 $\frac{dy}{dx} + 2y = \sin x$.This is a linear differential equation of the form $\frac{dy}{dx} + Py = Q$,where

Vedclass · Accepted Answer

$y = \frac{1}{5}(2 \sin x - \cos x) + Ce^{-2x}$

Vedclass · Answer

(A) The given differential equation is $\frac{dy}{dx} + 2y = \sin x$.This is a linear differential equation of the form $\frac{dy}{dx} + Py = Q$,where $P = 2$ and $Q = \sin x$.First,we find the integrating factor $(I.F.)$:$I.F. = e^{\int P dx} = e^{\int 2 dx} = e^{2x}$.The general solution is given by $y(I.F.) = \int (Q 	imes I.F.) dx + C$.Substituting the values,we get $y e^{2x} = \int \sin x \cdot e^{2x} dx + C$  ..........$(1)$.To solve $I = \int e^{2x} \sin x dx$,we use integration by parts:$I = \sin x \cdot \frac{e^{2x}}{2} - \int \cos x \cdot \frac{e^{2x}}{2} dx = \frac{e^{2x} \sin x}{2} - \frac{1}{2} \int e^{2x} \cos x dx$.Applying integration by parts again to $\int e^{2x} \cos x dx$:$I = \frac{e^{2x} \sin x}{2} - \frac{1}{2} \left[ \cos x \cdot \frac{e^{2x}}{2} - \int (-\sin x) \cdot \frac{e^{2x}}{2} dx ight] = \frac{e^{2x} \sin x}{2} - \frac{e^{2x} \cos x}{4} - \frac{1}{4} I$.$I + \frac{1}{4} I = \frac{e^{2x}}{4} (2 \sin x - \cos x) \Rightarrow \frac{5}{4} I = \frac{e^{2x}}{4} (2 \sin x - \cos x) \Rightarrow I = \frac{e^{2x}}{5} (2 \sin x - \cos x)$.Substituting this back into equation $(1)$:$y e^{2x} = \frac{e^{2x}}{5} (2 \sin x - \cos x) + C$.Dividing by $e^{2x}$,we get $y = \frac{1}{5} (2 \sin x - \cos x) + Ce^{-2x}$.

Find the general solution of the differential equation: $\frac{dy}{dx} + 2y = \sin x$.

Similar Questions

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

Let $y=y(x)$ be the solution of the differential equation $(x^2+1) y^{\prime}-2 x y=(x^4+2 x^2+1) \cos x$,with $y(0)=1$. Then $\int_{-3}^3 y(x) d x$ is :

Let $y = y(x)$ be the solution of the differential equation $\frac{dy}{dx} + 2y = f(x)$,where $f(x) = \begin{cases} 1, & x \in [0, 1] \\ 0, & \text{otherwise} \end{cases}$. If $y(0) = 0$,then $y\left(\frac{3}{2}\right)$ 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: $[y(1) = 0]$

Vedclass Test Series

Exam Paper Generator

Online Exam Module