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

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

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

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

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

Solution steps

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

Solve linear ODE:

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

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

y^{^{''}} - y = t - 2, y (2) = 3, y^{^{'}} (2) = 0

y^{^{''}} - y = sin (t)

y^{^{''}} + y^{^{'}} - 2 y = x + sin (2 x), y (0) = 1, y^{^{'}} (0) = 0

4 y^{^{''}} - 16 y^{^{'}} + 64 y = - \frac{2 e ^{8 x}}{3 x}

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

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