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

The solution for y^{''}-4y^'=sin(3x) is y=c_{1}e^{4x}+c_{2}-1/25 sin(3x)+4/75 cos(3x)

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

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

y = c_{1} e^{4 x} + c_{2} - \frac{1}{25} sin (3 x) + \frac{4}{75} cos (3 x)

Solution steps

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

Solve linear ODE:

y = c_{1} e^{4 x} + c_{2} - \frac{1}{25} sin (3 x) + \frac{4}{75} cos (3 x)

y = c_{1} e^{4 x} + c_{2} - \frac{1}{25} sin (3 x) + \frac{4}{75} cos (3 x)

7 x^{6} - 4 x^{5}

\int cos (2 x) + sin (x) d x

\int 10000 - x^{2} d x

y = x^{2} - 4 x, y = - 2 x

x \to 2 lim (- 2 x^{2})

What is the solution for y^{''}-4y^'=sin(3x) ?
The solution for y^{''}-4y^'=sin(3x) is y=c_{1}e^{4x}+c_{2}-1/25 sin(3x)+4/75 cos(3x)