What is the solution for y^{''}-3y^'+2y=e^{3x}cos(e^x) ?

The solution for y^{''}-3y^'+2y=e^{3x}cos(e^x) is y=c_{1}e^{2x}+c_{2}e^x-e^xcos(e^x)

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y^{''}-3y^'+2y=e^{3x}cos(e^x)

y^{^{''}} - 3 y^{^{'}} + 2 y = e^{3 x} cos (e^{x})

y = c_{1} e^{2 x} + c_{2} e^{x} - e^{x} cos (e^{x})

Solution steps

y^{''} (x) - 3 y^{'} (x) + 2 y = e^{3 x} cos (e^{x})

Solve linear ODE:

y = c_{1} e^{2 x} + c_{2} e^{x} - e^{x} cos (e^{x})

y = c_{1} e^{2 x} + c_{2} e^{x} - e^{x} cos (e^{x})

\frac{d ^{2} y}{d x ^{2}} - 9 \frac{d y}{d x} + 2 y = x e^{x}

(d^{2} - 1) y = x e^{x} sin (x)

y^{^{''}} + 11 y^{^{'}} + 18 y = 6 x

y^{^{''}} - 2 y^{^{'}} + y = x e^{- x} + cos (x)

y^{^{''}} - y^{^{'}} = - 8

What is the solution for y^{''}-3y^'+2y=e^{3x}cos(e^x) ?
The solution for y^{''}-3y^'+2y=e^{3x}cos(e^x) is y=c_{1}e^{2x}+c_{2}e^x-e^xcos(e^x)