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Homework Statement
Solve the DE:
y''-2y'+2y=0
and express your answer in the form
ce[tex]\alpha[/tex]x*sin([tex]\beta[/tex]x+[tex]\gamma[/tex])
where alpha = ;
and beta =
Homework Equations
y = e[tex]\lambda[/tex]x
The Attempt at a Solution
When you plug in the values of y into the DE and solve for lambda you get
[tex]\lambda[/tex] = 1+ i or 1-i
A particular solution to the DE is
y(x)= e(1+i)x
when this is expanded using Euler's formula to solve the general solution to the DE, I get
y(x) = k1excos(x)+k2exsin(x)
When looking at the form of the solution they want me to express my answer in I feel like that are asking for
k2exsin(x) = ce[tex]\alpha[/tex]x*sin([tex]\beta[/tex]x+[tex]\gamma[/tex])
alpha = 1 but how do I get beta? I don't think I can express my answer with gamma
Thanks!