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}}

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)}