What is the solution for y^{''}+2y^'=4,y(0)=3,y^'(0)=5 ?

The solution for y^{''}+2y^'=4,y(0)=3,y^'(0)=5 is y=3+3/2-3/2 e^{-2t}+2t

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y^{''}+2y^'=4,y(0)=3,y^'(0)=5

y^{^{''}} + 2 y^{^{'}} = 4, y (0) = 3, y^{^{'}} (0) = 5

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

Solution steps

y^{''} (t) + 2 y^{'} (t) = 4

Solve linear ODE:

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

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

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

y^{^{''}} + 3 y^{^{'}} + 2 y = x^{2} - 1

y^{^{''}} - 3 y^{^{'}} - 4 y = 3 e^{2} x

y^{^{''}} - 4 y^{^{'}} + 5 y = \frac{2 e ^{2 x}}{sin ( x )}

3 y^{^{''}} - 6 y^{^{'}} + 6 y = e^{x} \cdot sec (x)

What is the solution for y^{''}+2y^'=4,y(0)=3,y^'(0)=5 ?
The solution for y^{''}+2y^'=4,y(0)=3,y^'(0)=5 is y=3+3/2-3/2 e^{-2t}+2t