What is the solution for y^{''}+y^'=t,y(0)=1,y^'(0)=2 ?

The solution for y^{''}+y^'=t,y(0)=1,y^'(0)=2 is y=(t^2)/2-t-3e^{-t}+4

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y^{''}+y^'=t,y(0)=1,y^'(0)=2

y^{^{''}} + y^{^{'}} = t, y (0) = 1, y^{^{'}} (0) = 2

y = \frac{t ^{2}}{2} - t - 3 e^{- t} + 4

Solution steps

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

Solve linear ODE:

y = \frac{t ^{2}}{2} - t - 3 e^{- t} + 4

y = \frac{t ^{2}}{2} - t - 3 e^{- t} + 4

y^{^{''''}} + 2 y^{^{''}} + y = 5 + cos (4 t)

4 y^{^{''}} - 16 y = 1

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

y^{^{''}} + y = t, y (0) = 1, y^{^{'}} (0) = - 2

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

What is the solution for y^{''}+y^'=t,y(0)=1,y^'(0)=2 ?
The solution for y^{''}+y^'=t,y(0)=1,y^'(0)=2 is y=(t^2)/2-t-3e^{-t}+4