Solution Found!
Answer: In 19–22 solve each differential equation by
Chapter 4, Problem 22E(choose chapter or problem)
In Problems 19–22 solve each differential equation by variation of parameters, subject to the initial conditions y(0) = 1, \(y^{\prime}(0)=0\).
\(y^{\prime \prime}-4 y^{\prime}+4 y=\left(12 x^{2}-6 x\right) e^{2 x}\)
Text Transcription:
y^prime 0=0
y^prime prime-4 y^prime+4 y=left12 x^2-6 x\right e^2 x
Questions & Answers
QUESTION:
In Problems 19–22 solve each differential equation by variation of parameters, subject to the initial conditions y(0) = 1, \(y^{\prime}(0)=0\).
\(y^{\prime \prime}-4 y^{\prime}+4 y=\left(12 x^{2}-6 x\right) e^{2 x}\)
Text Transcription:
y^prime 0=0
y^prime prime-4 y^prime+4 y=left12 x^2-6 x\right e^2 x
ANSWER:Step 1 of 6
Given that
We have to solve the differential equation by variation of parameters, subject to
the initial conditions ,