Answer
The actual result: $23.146$
The approximation $25.958$
Difference between the result and approximation: $25.958-23.146=2.812$
Work Step by Step
We are given $\frac{dy}{dx}-2y=e^{2x}$
Using Euler's method $g(x,y)=e^{2x}+2y$
Since $x=0, y=10$
$g(x_{0},y_{0})=21$ and $y_{1}=y_{0}+g(x_{0},y_{0})h=10+21\times0.1=12.1$
Now $x_{1}=0.1, y_{1}=12.1$
and $g(x_{1},y_{1})=25.421$
Then $y_{2}=12.1+(25.421)\times0.1=14.642$
$y_{3}=14.642+(30.776)\times0.1=17.72$
$y_{4}=17.72+(37.262)\times0.1=21.446$
$y_{5}=21.446+(45.118)\times0.1=25.958$ $y(0.4)=y_{5}=25.958$
$\frac{dy}{dx}-2y=e^{2x}$
This equation is written in the form $\frac{dy}{dx}+P(x)y=Q(x)$ we can note that $P(x)=-2$
The integrating factor is $I(x)=e^{\int P(x)dx}=e^{\int-2dx}=e^{-2x}$
multiplying both sides of the differential equation by $e^{-2x}$
$e^{-2x}\frac{dy}{dx}-2e^{-2x}y=1$
Write the terms on the left in the form $D_{x}[I(x)y]$ $D_{x}[e^{-2x}y]=1$ solve for y integrating both sides $e^{-2x}y=\int 1dx$
$e^{-2x}y=x+C$
$y=\frac{x+C}{e^{-2x}}$
Find the particular solution by substituting 0 for x and 10 for y $C=10$
$y=\frac{x+10}{e^{-2x}}$
$\rightarrow y=23.146$
Difference between the result and approximation: $25.958-23.146=2.812$
