What is the solution for 2y^'+10y=3u,y(0)=-2 ?

The solution for 2y^'+10y=3u,y(0)=-2 is y=-(3ue^{-5t})/(10)-2e^{-5t}+(3u)/(10)

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2y^'+10y=3u,y(0)=-2

2 y^{^{'}} + 10 y = 3 u, y (0) = - 2

y = - \frac{3 u e ^{- 5 t}}{10} - 2 e^{- 5 t} + \frac{3 u}{10}

Solution steps

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

Solve separable ODE:

y = - \frac{3 u e ^{- 5 t}}{10} - 2 e^{- 5 t} + \frac{3 u}{10}

y = - \frac{3 u e ^{- 5 t}}{10} - 2 e^{- 5 t} + \frac{3 u}{10}

2 y^{^{'}} + y = 0, y (0) = e^{\frac{- x}{2}}

y^{^{'}} = \frac{x ^{2} - 6}{y ^{2}}, y (0) = 2

\frac{d y}{d x} = \frac{x ^{2}}{x ^{2} ( 3 y ^{2} + 1 )}, y (1) = 4

\frac{d y}{d x} = x^{5} y^{2}

5 x y y^{^{'}} = 1

What is the solution for 2y^'+10y=3u,y(0)=-2 ?
The solution for 2y^'+10y=3u,y(0)=-2 is y=-(3ue^{-5t})/(10)-2e^{-5t}+(3u)/(10)