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

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

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

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

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

Solution steps

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

Solve linear ODE:

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

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

y^{^{''}} - 2 y^{^{'}} + y = \frac{e ^{x}}{x ^{5}}

y^{^{''}} - 4 y = 3 x e^{2 x}

y^{^{''}} - 25 y = 3 x^{2} - 2 + e^{x}

y^{^{'''}} - y^{^{''}} = x

y^{^{''}} - y = 2 e^{x} + 2 e^{- x} + x

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