What is the solution for y^{''}-3y^'=sin(4t) ?

The solution for y^{''}-3y^'=sin(4t) is y=c_{1}e^{3t}+c_{2}-1/25 sin(4t)+3/100 cos(4t)

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y^{''}-3y^'=sin(4t)

y^{^{''}} - 3 y^{^{'}} = sin (4 t)

y = c_{1} e^{3 t} + c_{2} - \frac{1}{25} sin (4 t) + \frac{3}{100} cos (4 t)

Solution steps

y^{''} (t) - 3 y^{'} (t) = sin (4 t)

Solve linear ODE:

y = c_{1} e^{3 t} + c_{2} - \frac{1}{25} sin (4 t) + \frac{3}{100} cos (4 t)

y = c_{1} e^{3 t} + c_{2} - \frac{1}{25} sin (4 t) + \frac{3}{100} cos (4 t)

y^{^{''}} + 4 y = 2 x, y (0) = 0, y^{^{'}} (0) = 2

y^{^{''}} + a y^{^{'}} + b y = x, y (0) = 0, y^{^{'}} (0) = 0

(D^{2} - 4 D + 3) y = 20 cos (x)

y^{^{''}} + (- 4) y^{^{'}} + 4 y = \frac{2 e ^{2}}{x}

y^{^{''}} - 9 y = 20 sin (x) + 3 x^{2}

What is the solution for y^{''}-3y^'=sin(4t) ?
The solution for y^{''}-3y^'=sin(4t) is y=c_{1}e^{3t}+c_{2}-1/25 sin(4t)+3/100 cos(4t)