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

The solution for y^{''}+2y^'-3y=2cos(x) is y=c_{1}e^x+c_{2}e^{-3x}+1/5 sin(x)-2/5 cos(x)

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

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

y = c_{1} e^{x} + c_{2} e^{- 3 x} + \frac{1}{5} sin (x) - \frac{2}{5} cos (x)

Solution steps

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

Solve linear ODE:

y = c_{1} e^{x} + c_{2} e^{- 3 x} + \frac{1}{5} sin (x) - \frac{2}{5} cos (x)

y = c_{1} e^{x} + c_{2} e^{- 3 x} + \frac{1}{5} sin (x) - \frac{2}{5} cos (x)

y^{^{''}} - 4 y = 1

y^{^{''}} - 4 y = 5

\frac{d ^{2} y}{d x ^{2}} - 2 \frac{d y}{d x} + y = x^{2} + 1

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

y^{^{''}} + 2 y^{^{'}} + 5 y = 2 cos (x)

What is the solution for y^{''}+2y^'-3y=2cos(x) ?
The solution for y^{''}+2y^'-3y=2cos(x) is y=c_{1}e^x+c_{2}e^{-3x}+1/5 sin(x)-2/5 cos(x)