Question

Solve the given initial-value problem. y'' 4y' 5y = 35e−4x, y(0) = −5, y'(0) = 1

Answer 1

The solution for the initial value problem is $y_{g} = e^{-2x} (-12cos(x) + 5sin(x)) + 7e^{-4x}$

Given,

y" + 4y' + 5y = 35$e^{-4x}$

y(0) = -5

y'(0) = 1

Solve this homogenous equation to get $y_{h}$

According to differential operator theorem,

$y_{h}$ = $e^{ax}$( A cos (bx) + B sin (bx)), where A and B are constants.

Therefore,

y" + 4y' + 5y = 0

($D^{2}$ + 4D + 5)y = 0

D = -2± i    

$y_{h} = e^{-2x} ( A cos (x) + B sin (x))$

Now, solve for $y_{p}$

A function of the kind $ce^{-4x}$ is the function on the right, we are trying a solution of the form $y_{p} =ce^{-4x}$, here c is a constant.

$y_{p} " + 4y_{p} ' + 5y_{p} = 35e^{-4x} \\=16ce^{-4x} -16ce^{-4x} +5ce^{-4x} = 35e^{-4x} \\= 5ce^{-4x} =35e^{-4x} \\c=\frac{35}{5} =7\\y_{p} =7e^{-4x}$

Then the general solution will be like:

$y_{g} =y_{h} +y_{p}$

    = $e^{-2x} (Acos(x)+Bsin(x))+7e^{-4x}$

$y_{g}(0)=-5=A+7=-12\\y_{g} '(0)=e^{-2x} (-Asin(x)+Bcos(x))-2e^{-2x} (Acos(x)+Bsin(x))-28e^{-4x} \\y'_{g} (0)=1=B-2A-28$

       B = 1 - 24 + 28 = 5

Then the solution for the given initial value problem is

$y_{g} =e^{-2x} (-12cos(x)+5sin(x))+7e^{-4x}$

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