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Sagot :
The solution for the initial value problem is [tex]y_{g} = e^{-2x} (-12cos(x) + 5sin(x)) + 7e^{-4x}[/tex]
Given,
y" + 4y' + 5y = 35[tex]e^{-4x}[/tex]
y(0) = -5
y'(0) = 1
Solve this homogenous equation to get [tex]y_{h}[/tex]
According to differential operator theorem,
[tex]y_{h}[/tex] = [tex]e^{ax}[/tex]( A cos (bx) + B sin (bx)), where A and B are constants.
Therefore,
y" + 4y' + 5y = 0
([tex]D^{2}[/tex] + 4D + 5)y = 0
D = -2± i
[tex]y_{h} = e^{-2x} ( A cos (x) + B sin (x))[/tex]
Now, solve for [tex]y_{p}[/tex]
A function of the kind [tex]ce^{-4x}[/tex] is the function on the right, we are trying a solution of the form [tex]y_{p} =ce^{-4x}[/tex], here c is a constant.
[tex]y_{p} " + 4y_{p} ' + 5y_{p} = 35e^{-4x} \\=16ce^{-4x} -16ce^{-4x} +5ce^{-4x} = 35e^{-4x} \\= 5ce^{-4x} =35e^{-4x} \\c=\frac{35}{5} =7\\y_{p} =7e^{-4x}[/tex]
Then the general solution will be like:
[tex]y_{g} =y_{h} +y_{p} \\[/tex]
= [tex]e^{-2x} (Acos(x)+Bsin(x))+7e^{-4x}[/tex]
[tex]y_{g}(0)=-5=A+7=-12\\y_{g} '(0)=e^{-2x} (-Asin(x)+Bcos(x))-2e^{-2x} (Acos(x)+Bsin(x))-28e^{-4x} \\y'_{g} (0)=1=B-2A-28\\[/tex]
B = 1 - 24 + 28 = 5
Then the solution for the given initial value problem is
[tex]y_{g} =e^{-2x} (-12cos(x)+5sin(x))+7e^{-4x}[/tex]
Learn more about initial value problem here: https://brainly.com/question/8599681
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