## Calculus with Applications (10th Edition)

The actual result: $23.146$ The approximation $25.958$ Difference between the result and approximation: $25.958-23.146=2.812$
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: $25.958-23.146=2.812$