Undetermined Coefficients/Var. of Parameters: y'' - y' = e^x

[paulxzt]

y'' - y' = e^x

a) solve using undetermined coefficient
b) solve using variation of parameters

a)
auxiliary eq: r^2 - r = 0, r = 0, 1

yc(x) = C1e^x + C2

yp(x) = xAe^x

I figured out y' and y'' and plugged it in and solved for A to get A = 1 and
y(x) = C1e^x + C2 + xe^x.

am i doing this right?
can someone help me get started on VAP
ho

 

[soroban]

Re: Undetermined Coefficients/Var. of Parameters: y'' - y' =

Hello, paulxzt!

$\displaystyle y''\,-\,y' \:= \:e^x$

a) Solve using undetermined coefficient
b) Solve using variation of parameters


a) auxiliary eq: $\displaystyle \:r^2\, -\,r \:= \:0\;\;\Rightarrow\;\;r \:= \:0,\,1\;\;\;\Rightarrow\;\;\;y_c(x) \:= \:C_1e^x\,+\,C_2$

$\displaystyle y_p(x) \:= \:xAe^x$
I figured out $\displaystyle y'$ and $\displaystyle y''$ and plugged it in, and solved for $\displaystyle A$ to get: $\displaystyle \,A \,=\, 1$

Therefore: $\displaystyle \:y(x) \:= \:C_1e^x\,+\,C_2\,+\,xe^x$

Am i doing this right? . . . . Yes! . Good work!

Can someone help me get started on VAP?

We have: \(\displaystyle \:y_c \;=\;C_1e^x\,+\,C_2\\)
. . where $\displaystyle C_1$ and $\displaystyle C_2$ are functions of $\displaystyle x.$

Differentiate: $\displaystyle \:y'\;=\;C_1e^x\,+\,C_1'e^x\,+\,C_2'\;\;\Rightarrow\;\;$$\displaystyle {\color{blue} C_1'e^x\,+\,C_2' \:=\:0}$

We have: $\displaystyle \:y' \:=\:C_1e^x$
Differentiate: $\displaystyle \:y'' \;=\;C_1e^x\,+\,C_1'e^x\;\;\Rightarrow\;\;$$\displaystyle {\color{blue}C_1'e^x\:=\:e^x}$

I assume you know the rest . . .
. . Solve the system of equation for $\displaystyle C_1'$ and $\displaystyle C_2'$
. . Determine $\displaystyle C_1$ and $\displaystyle C_2$.
. . Substitute into $\displaystyle y_c$