What is the solution for y^'+2y=10sin(t) ?

The solution for y^'+2y=10sin(t) is y=2(2sin(t)-cos(t))+(c_{1})/(e^{2t)}

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y^'+2y=10sin(t)

y^{^{'}} + 2 y = 10 sin (t)

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

Solution steps

y^{'} (t) + 2 y = 10 sin (t)

Solve linear ODE:

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

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

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

\frac{\partial}{\partial t} (cos (3 t))

\frac{x ^{2}}{2 ( 1 - x )}

f (x) = lo g_{6} (2 x^{2} + 4), at x = 4

(- 1, - 5), (1, - 2)

What is the solution for y^'+2y=10sin(t) ?
The solution for y^'+2y=10sin(t) is y=2(2sin(t)-cos(t))+(c_{1})/(e^{2t)}