What is the solution for y^{''}-3y^'=4sin(x) ?

The solution for y^{''}-3y^'=4sin(x) is y=c_{1}e^{3x}+c_{2}-2/5 sin(x)+6/5 cos(x)

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y^{''}-3y^'=4sin(x)

y^{^{''}} - 3 y^{^{'}} = 4 sin (x)

y = c_{1} e^{3 x} + c_{2} - \frac{2}{5} sin (x) + \frac{6}{5} cos (x)

Solution steps

y^{''} (x) - 3 y^{'} (x) = 4 sin (x)

Solve linear ODE:

y = c_{1} e^{3 x} + c_{2} - \frac{2}{5} sin (x) + \frac{6}{5} cos (x)

y = c_{1} e^{3 x} + c_{2} - \frac{2}{5} sin (x) + \frac{6}{5} cos (x)

y^{^{''}} - y^{^{'}} + y = t^{2} + cos (t)

y^{^{''}} - 2 y^{^{'}} - 3 y = 6 x e^{3 x}

y^{^{''}} + 4 y^{^{'}} + 3 y = e^{5 x}

y^{^{'''}} - 6 y^{^{''}} + 11 y^{^{'}} - 6 y + x^{2} = - 1

x^{^{''}} + 5 x^{^{'}} + 4 x = 12.5 cos (4 t)

What is the solution for y^{''}-3y^'=4sin(x) ?
The solution for y^{''}-3y^'=4sin(x) is y=c_{1}e^{3x}+c_{2}-2/5 sin(x)+6/5 cos(x)