What is the solution for (d^2x)/(dt^2)+4x=sin(1t),x(0)=1 ?

The solution for (d^2x)/(dt^2)+4x=sin(1t),x(0)=1 is x=cos(2t)+c_{2}sin(2t)+1/3 sin(t)

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(d^2x)/(dt^2)+4x=sin(1t),x(0)=1

\frac{d ^{2} x}{d t ^{2}} + 4 x = sin (1 t), x (0) = 1

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

Solution steps

\frac{d ^{2} x}{d t ^{2}} + 4 x = sin (1 t)

Solve linear ODE:

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

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

\frac{d}{d x} (- \frac{1}{15} cos (15 x))

\int e^{2 - x} d x

\int \frac{x - 1}{x} d x

y = \frac{3}{x}

\int_{0}^{1.2} 9 cos (1.9 \cdot x) d x

What is the solution for (d^2x)/(dt^2)+4x=sin(1t),x(0)=1 ?
The solution for (d^2x)/(dt^2)+4x=sin(1t),x(0)=1 is x=cos(2t)+c_{2}sin(2t)+1/3 sin(t)