Answer: In 19–22 solve each differential equation by

Chapter 4, Problem 22E

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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

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 ,

         

 

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