What is the solution for (d^2x)/(dt^2)+36x=10sin(4t),x(0)=0 ?

The solution for (d^2x)/(dt^2)+36x=10sin(4t),x(0)=0 is x=c_{2}sin(6t)+1/2 sin(4t)

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(d^2x)/(dt^2)+36x=10sin(4t),x(0)=0

\frac{d ^{2} x}{d t ^{2}} + 36 x = 10 sin (4 t), x (0) = 0

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

Solution steps

\frac{d ^{2} x}{d t ^{2}} + 36 x = 10 sin (4 t)

Solve linear ODE:

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

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

y^{^{''}} - 3 y^{^{'}} = 3 e^{3 t}

y^{^{''}} + 16 y = 10 cos (4 x) + 15 sin (4 x)

y^{^{''}} - 10 y^{^{'}} + 24 y = 48 x^{2} + 10

y^{^{''}} + 4 y = t^{2} + 10 e^{t}, y (0) = 0, y^{^{'}} (0) = 3

y^{^{''}} + y = 4 x \cdot sin (x)

What is the solution for (d^2x)/(dt^2)+36x=10sin(4t),x(0)=0 ?
The solution for (d^2x)/(dt^2)+36x=10sin(4t),x(0)=0 is x=c_{2}sin(6t)+1/2 sin(4t)