What is the solution for y^'+2y=3cos(t) ?

The solution for y^'+2y=3cos(t) is y=(3(sin(t)+2cos(t)))/5+(c_{1})/(e^{2t)}

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y^'+2y=3cos(t)

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

y = \frac{3 ( sin ( t ) + 2 cos ( t ) )}{5} + \frac{c _{1}}{e ^{2 t}}

Solution steps

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

Solve linear ODE:

y = \frac{3 ( sin ( t ) + 2 cos ( t ) )}{5} + \frac{c _{1}}{e ^{2 t}}

y = \frac{3 ( sin ( t ) + 2 cos ( t ) )}{5} + \frac{c _{1}}{e ^{2 t}}

\int (2 - sin (x))^{2} d x

x \to 0 + lim (x \cdot ln (x))

5

6 x, 8 x^{2}

What is the solution for y^'+2y=3cos(t) ?
The solution for y^'+2y=3cos(t) is y=(3(sin(t)+2cos(t)))/5+(c_{1})/(e^{2t)}