Gregory Emery

2021-12-31

Find the differential equations of the following. Show complete solution. $\frac{dy}{dx}+y={e}^{2x}$

### Answer & Explanation

Janet Young

Step 1
Given: $\frac{dy}{dx}+y={e}^{2x}$
To Find: To find the general solution of the given differential equation.
Step 2
Solution:
$\frac{dy}{dx}+y={e}^{2x}$
$⇒\frac{dy}{dx}+1y={e}^{2x}$ (1)
We observe that equation (1) is a first order linear ordinary differential $=q$
equation of the form, $\frac{dy}{dx}+py$
Where,

Multiplying equation (1) by the Integrating factor $={e}^{x}$, we get
${e}^{x}\frac{dy}{dx}+{e}^{x}y={e}^{x}\left({e}^{2x}\right)$
${e}^{x}\frac{dy}{dx}+{e}^{x}y={e}^{x+2x}$
$⇒{e}^{x}\frac{dy}{dx}+{e}^{x}y={e}^{3x}$
$⇒\frac{d}{dx}\left(y.{e}^{x}\right)={e}^{3x}$
$⇒d\left(y.{e}^{x}\right)={e}^{3x}dx$
Integrating both the sides we get
$\int d\left(y.{e}^{x}\right)=\int {e}^{3x}dx$
$⇒y.{e}^{x}=\left(\frac{{e}^{3x}}{3}\right)+c$, where $c=$ constant of integration
$⇒y.{e}^{x}=\frac{1}{3}{e}^{3x}+c$
$⇒y=\frac{1}{{e}^{x}}\left(\frac{1}{3}{e}^{3x}+c\right)$
$⇒y=\frac{1}{3}\left(\frac{{e}^{3x}}{{e}^{x}}\right)+\frac{c}{{e}^{x}}$<

Hector Roberts

Given $\frac{dy}{dx}+y={e}^{2x}$
It is in the linear differential equation form of $\frac{dy}{dx}+Py=Q$
Here,
We first find the integrating factor(IF)
$I={e}^{\int pdx}={e}^{\int 1.dx}={e}^{x}$
Then we multiply the differential equation with IF to get
$y×IF=\int Q×IF.dx+C$
$⇒{e}^{x}y={e}^{x}\left({e}^{2x}\right)+C$
$⇒{e}^{x}\left(y\right)=\frac{{e}^{3x}}{3}+C$
$⇒y=\frac{{e}^{3x}}{3}{e}^{x}+c.{e}^{-x}$
General solution, $y=\frac{{e}^{2x}}{3}+c.{e}^{-x}$

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