What is the solution for 2y^{''}+y^'=4x,y(0)=7,y^'(0)=-10 ?

The solution for 2y^{''}+y^'=4x,y(0)=7,y^'(0)=-10 is y=2x^2-8x+4e^{-x/2}+3

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2y^{''}+y^'=4x,y(0)=7,y^'(0)=-10

2 y^{^{''}} + y^{^{'}} = 4 x, y (0) = 7, y^{^{'}} (0) = - 10

y = 2 x^{2} - 8 x + 4 e^{- \frac{x}{2}} + 3

Solution steps

2 y^{''} (x) + y^{'} (x) = 4 x

Solve linear ODE:

y = 2 x^{2} - 8 x + 4 e^{- \frac{x}{2}} + 3

y = 2 x^{2} - 8 x + 4 e^{- \frac{x}{2}} + 3

y^{^{''}} + 4 y = r (x)

y^{^{''}} - 2 y^{^{'}} + y = x^{3} + 5 x

y^{^{''}} - 2 y^{^{'}} + y = \frac{e ^{x}}{x} + e^{x} x^{2} + 5 cos (2 x)

y^{^{''}} - 2 y^{^{'}} + 2 y = 2 x - 2, y^{^{'}} (π) = π, y (0) = 0

\frac{d ^{2} y}{d x ^{2}} + 4 y = sin^{2} (x)

What is the solution for 2y^{''}+y^'=4x,y(0)=7,y^'(0)=-10 ?
The solution for 2y^{''}+y^'=4x,y(0)=7,y^'(0)=-10 is y=2x^2-8x+4e^{-x/2}+3