What is the solution for y^{''}+4y=3sin(2t) ?

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

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y^{''}+4y=3sin(2t)

y^{^{''}} + 4 y = 3 sin (2 t)

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

Solution steps

y^{''} (t) + 4 y = 3 sin (2 t)

Solve linear ODE:

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

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

5^{x^{5}}

\int \frac{1}{ln ( 2 x )} d x

2 y = 5 x, y = 3, 2 y + 2 x = 7

\int x (3^{x}) d x

\int (\frac{1}{- 0.4 t ^{3}}) d t

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