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

The solution for y^{''}-3y^'+2y=e^x+cos(x) is y=c_{1}e^{2x}+c_{2}e^x-e^xx-3/10 sin(x)+1/10 cos(x)

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

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

y = c_{1} e^{2 x} + c_{2} e^{x} - e^{x} x - \frac{3}{10} sin (x) + \frac{1}{10} cos (x)

Solution steps

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

Solve linear ODE:

y = c_{1} e^{2 x} + c_{2} e^{x} - e^{x} x - \frac{3}{10} sin (x) + \frac{1}{10} cos (x)

y = c_{1} e^{2 x} + c_{2} e^{x} - e^{x} x - \frac{3}{10} sin (x) + \frac{1}{10} cos (x)

y^{^{''}} - 2 y^{^{'}} - 3 y = 8 e x - 12

\frac{d ^{2} y}{d x ^{2}} + \frac{d y}{d x} - 6 y = 18

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

y^{^{''}} + 49 y = sec (7 x)

\frac{d ^{2} y}{d x ^{2}} + 4 y = cos (2 x) + x e^{3 x}

What is the solution for y^{''}-3y^'+2y=e^x+cos(x) ?
The solution for y^{''}-3y^'+2y=e^x+cos(x) is y=c_{1}e^{2x}+c_{2}e^x-e^xx-3/10 sin(x)+1/10 cos(x)