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

paulxzt

Junior Member
Joined
Aug 30, 2006
Messages
65
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
 
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 \(\displaystyle y_c\)

 
Top