What is the solution for 2y^{''}+3y^'-5y=2cos(x) ?

The solution for 2y^{''}+3y^'-5y=2cos(x) is y=c_{1}e^x+c_{2}e^{-(5x)/2}+3/29 sin(x)-7/29 cos(x)

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2y^{''}+3y^'-5y=2cos(x)

2 y^{^{''}} + 3 y^{^{'}} - 5 y = 2 cos (x)

y = c_{1} e^{x} + c_{2} e^{- \frac{5 x}{2}} + \frac{3}{29} sin (x) - \frac{7}{29} cos (x)

Solution steps

2 y^{''} (x) + 3 y^{'} (x) - 5 y = 2 cos (x)

Solve linear ODE:

y = c_{1} e^{x} + c_{2} e^{- \frac{5 x}{2}} + \frac{3}{29} sin (x) - \frac{7}{29} cos (x)

y = c_{1} e^{x} + c_{2} e^{- \frac{5 x}{2}} + \frac{3}{29} sin (x) - \frac{7}{29} cos (x)

What is the solution for 2y^{''}+3y^'-5y=2cos(x) ?
The solution for 2y^{''}+3y^'-5y=2cos(x) is y=c_{1}e^x+c_{2}e^{-(5x)/2}+3/29 sin(x)-7/29 cos(x)