What is the solution for (d^2x)/(dt^2)+36x=5sin(4t),x(0)=0 ?

The solution for (d^2x)/(dt^2)+36x=5sin(4t),x(0)=0 is x=c_{2}sin(6t)+1/4 sin(4t)

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(d^2x)/(dt^2)+36x=5sin(4t),x(0)=0

\frac{d ^{2} x}{d t ^{2}} + 36 x = 5 sin (4 t), x (0) = 0

x = c_{2} sin (6 t) + \frac{1}{4} sin (4 t)

Solution steps

\frac{d ^{2} x}{d t ^{2}} + 36 x = 5 sin (4 t)

Solve linear ODE:

x = c_{2} sin (6 t) + \frac{1}{4} sin (4 t)

x = c_{2} sin (6 t) + \frac{1}{4} sin (4 t)

y^{^{''}} - 2 y^{^{'}} - 3 y = x^{2}

x y^{^{'''}} - x y^{^{'}} = x^{4}

y^{^{''}} - y^{^{'}} = 5 e^{x}

\frac{d ^{2} y}{d x ^{2}} - 2 \frac{d y}{d x} - 3 y = 2 e^{3 x}

\frac{d ^{2} y}{d x ^{2}} - 4 \frac{d y}{d x} - 32 y = (x + 1) e^{x}

What is the solution for (d^2x)/(dt^2)+36x=5sin(4t),x(0)=0 ?
The solution for (d^2x)/(dt^2)+36x=5sin(4t),x(0)=0 is x=c_{2}sin(6t)+1/4 sin(4t)