What is the solution for y^{''}+y^'=1,y^1(0)=1 ?

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

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y^{''}+y^'=1,y^1(0)=1

y^{^{''}} + y^{^{'}} = 1, y^{1} (0) = 1

y = c_{1} + c_{2} e^{- t} + t

Solution steps

y^{''} (t) + y^{'} (t) = 1

Solve linear ODE:

y = c_{1} + c_{2} e^{- t} + t

y = c_{1} + c_{2} e^{- t} + t

y^{^{''}} + 2 y^{^{'}} - 3 y = sin (t)

(d^{2} - 2 d + 2) y = x^{2} e^{3} x

y^{^{''}} + y^{^{'}} - 2 y = 8 sin (2 x)

(d^{2} + 2 d + 1) y = cos^{2} (x)

(d^{2} + 3 d + 2) y = e - x + cos (x)

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