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

The solution for y^{''}+y^'=x,y(0)=1,y^'(0)=0 is y=(x^2)/2-x-e^{-x}+2

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

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

y = \frac{x ^{2}}{2} - x - e^{- x} + 2

Solution steps

y^{''} (x) + y^{'} (x) = x

Solve linear ODE:

y = \frac{x ^{2}}{2} - x - e^{- x} + 2

y = \frac{x ^{2}}{2} - x - e^{- x} + 2

y^{^{''}} - 12 y^{^{'}} + 36 y = 42 x + 2

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

y^{^{''}} + 2 y^{^{'}} - 3 y = e^{3 x} (2 + 8 x - 12 x^{2})

y^{^{''}} + 20 y^{^{'}} + 125 y = 9 sin (5 x)

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

What is the solution for y^{''}+y^'=x,y(0)=1,y^'(0)=0 ?
The solution for y^{''}+y^'=x,y(0)=1,y^'(0)=0 is y=(x^2)/2-x-e^{-x}+2