What is the solution for y^{''}+3y^'=10sin(t) ?

The solution for y^{''}+3y^'=10sin(t) is y=c_{1}+c_{2}e^{-3t}-sin(t)-3cos(t)

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y^{''}+3y^'=10sin(t)

y^{^{''}} + 3 y^{^{'}} = 10 sin (t)

y = c_{1} + c_{2} e^{- 3 t} - sin (t) - 3 cos (t)

Solution steps

y^{''} (t) + 3 y^{'} (t) = 10 sin (t)

Solve linear ODE:

y = c_{1} + c_{2} e^{- 3 t} - sin (t) - 3 cos (t)

y = c_{1} + c_{2} e^{- 3 t} - sin (t) - 3 cos (t)

y = 5 x - 4 x, (1, 1)

\int_{0}^{1} (∣ 3 x - 1 ∣ - 2 ∣ x ∣) d x

- \frac{5}{4} t^{4} + 120 t^{2}

(x^{2} + 1) \cdot e^{x - 1}

\int x e^{- x y} d y

What is the solution for y^{''}+3y^'=10sin(t) ?
The solution for y^{''}+3y^'=10sin(t) is y=c_{1}+c_{2}e^{-3t}-sin(t)-3cos(t)