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

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

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

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

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

Solution steps

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

Solve linear ODE:

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

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

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