What is the solution for (dy)/(dt)*1/4+y=cos(t) ?

The solution for (dy)/(dt)*1/4+y=cos(t) is y=(4(sin(t)+4cos(t)))/(17)+(c_{1})/(e^{4t)}

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(dy)/(dt)*1/4+y=cos(t)

\frac{d y}{d t} \cdot \frac{1}{4} + y = cos (t)

y = \frac{4 ( sin ( t ) + 4 cos ( t ) )}{17} + \frac{c _{1}}{e ^{4 t}}

Solution steps

\frac{d y}{d t} \cdot \frac{1}{4} + y = cos (t)

Solve linear ODE:

y = \frac{4 ( sin ( t ) + 4 cos ( t ) )}{17} + \frac{c _{1}}{e ^{4 t}}

y = \frac{4 ( sin ( t ) + 4 cos ( t ) )}{17} + \frac{c _{1}}{e ^{4 t}}

\int - \frac{1}{3} e^{- 3 t} d t

\frac{d}{d x} (3 x + e^{x})

\frac{\partial}{\partial y} (4 - x^{2} - 7 y^{2})

2 y^{^{''}} + 3 y^{^{'}} + y = t^{2} + 2 sin (t)

\frac{d}{d x} (arcsin (\frac{2}{x}))

What is the solution for (dy)/(dt)*1/4+y=cos(t) ?
The solution for (dy)/(dt)*1/4+y=cos(t) is y=(4(sin(t)+4cos(t)))/(17)+(c_{1})/(e^{4t)}