What is the solution for (dx)/(dt)-5x=sin(2t),x(0)=pi ?

The solution for (dx)/(dt)-5x=sin(2t),x(0)=pi is x=-(2cos(2t))/(29)-(5sin(2t))/(29)+(pi+2/29)e^{5t}

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(dx)/(dt)-5x=sin(2t),x(0)=pi

\frac{d x}{d t} - 5 x = sin (2 t), x (0) = π

x = - \frac{2 cos ( 2 t )}{29} - \frac{5 sin ( 2 t )}{29} + (π + \frac{2}{29}) e^{5 t}

Solution steps

\frac{d x}{d t} - 5 x = sin (2 t)

Solve linear ODE:

x = - \frac{2 cos ( 2 t )}{29} - \frac{5 sin ( 2 t )}{29} + (π + \frac{2}{29}) e^{5 t}

x = - \frac{2 cos ( 2 t )}{29} - \frac{5 sin ( 2 t )}{29} + (π + \frac{2}{29}) e^{5 t}

\int tan^{5} (3 x) d x

\int \frac{4}{x ^{2} - 2 x} d x

5 sin (2 t)

s + 6

\frac{d}{d x} (\frac{1}{4 x})

What is the solution for (dx)/(dt)-5x=sin(2t),x(0)=pi ?
The solution for (dx)/(dt)-5x=sin(2t),x(0)=pi is x=-(2cos(2t))/(29)-(5sin(2t))/(29)+(pi+2/29)e^{5t}