What is the solution for y^'=-2y+5t,y(0)=-4 ?

The solution for y^'=-2y+5t,y(0)=-4 is y=(5(2t-1))/4-(11)/(4e^{2t)}

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y^'=-2y+5t,y(0)=-4

y^{^{'}} = - 2 y + 5 t, y (0) = - 4

y = \frac{5 ( 2 t - 1 )}{4} - \frac{11}{4 e ^{2 t}}

Solution steps

y^{'} (t) = - 2 y + 5 t

Solve linear ODE:

y = \frac{5 ( 2 t - 1 )}{4} - \frac{11}{4 e ^{2 t}}

y = \frac{5 ( 2 t - 1 )}{4} - \frac{11}{4 e ^{2 t}}

\int e^{9 x} cos (8 x) d x

\frac{\partial}{\partial z} (2 x^{\frac{y}{z}})

\int 4 x e^{- x} d x

y = 4 x

\frac{\partial}{\partial x} (x^{3} y^{2} - 4 y x^{2} + 3 y x - 4)

What is the solution for y^'=-2y+5t,y(0)=-4 ?
The solution for y^'=-2y+5t,y(0)=-4 is y=(5(2t-1))/4-(11)/(4e^{2t)}