What is the solution for (dx)/(dt)-4x=cos(5t) ?

The solution for (dx)/(dt)-4x=cos(5t) is x=(5sin(5t))/(41)-(4cos(5t))/(41)+c_{1}e^{4t}

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(dx)/(dt)-4x=cos(5t)

\frac{d x}{d t} - 4 x = cos (5 t)

x = \frac{5 sin ( 5 t )}{41} - \frac{4 cos ( 5 t )}{41} + c_{1} e^{4 t}

Solution steps

\frac{d x}{d t} - 4 x = cos (5 t)

Solve linear ODE:

x = \frac{5 sin ( 5 t )}{41} - \frac{4 cos ( 5 t )}{41} + c_{1} e^{4 t}

x = \frac{5 sin ( 5 t )}{41} - \frac{4 cos ( 5 t )}{41} + c_{1} e^{4 t}

f (x) = - 3 x^{3} - 2 x^{2} - x + 3, (-, 5)

n = 1 \sum \infty \frac{1}{n ^{7}}

\int sin (12 x) cos (9 x) d x

x \to 0 + lim (arctan (\frac{1}{x}))

\int_{- 2}^{0} (u^{5} - u^{3} + u^{2}) d u

What is the solution for (dx)/(dt)-4x=cos(5t) ?
The solution for (dx)/(dt)-4x=cos(5t) is x=(5sin(5t))/(41)-(4cos(5t))/(41)+c_{1}e^{4t}