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

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

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

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

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

Solution steps

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

Solve linear ODE:

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

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

\frac{d}{d n} (n^{n})

n = 0 \sum \infty n! (x - 5)^{n}

\int \frac{1}{x ( 1 + ln ( x ) ) ^{3}} d x

\frac{d r}{d θ} = \frac{r ^{2} + 2 r θ}{θ ^{2}}

x \to 0 lim (\frac{x}{sin ( 8 x )})

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