What is the solution for (d^2x)/(dt^2)+4x=cos(3t) ?

The solution for (d^2x)/(dt^2)+4x=cos(3t) is x=c_{1}cos(2t)+c_{2}sin(2t)-1/5 cos(3t)

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(d^2x)/(dt^2)+4x=cos(3t)

\frac{d ^{2} x}{d t ^{2}} + 4 x = cos (3 t)

x = c_{1} cos (2 t) + c_{2} sin (2 t) - \frac{1}{5} cos (3 t)

Solution steps

\frac{d ^{2} x}{d t ^{2}} + 4 x = cos (3 t)

Solve linear ODE:

x = c_{1} cos (2 t) + c_{2} sin (2 t) - \frac{1}{5} cos (3 t)

x = c_{1} cos (2 t) + c_{2} sin (2 t) - \frac{1}{5} cos (3 t)

x \to 2 lim (4 - a \cdot x)

n = 0 \sum \infty 3 x^{n}

\int_{0}^{1} \frac{1}{1 + x ^{\frac{1}{3}}} d x

\frac{d ^{2}}{d x ^{2}} (sin (x) + cos (x))

7 cos (π x), 8 x^{2} - 2, - 0.5, 0.5

What is the solution for (d^2x)/(dt^2)+4x=cos(3t) ?
The solution for (d^2x)/(dt^2)+4x=cos(3t) is x=c_{1}cos(2t)+c_{2}sin(2t)-1/5 cos(3t)