What is the solution for 4y^{''}-2y^'=4x,y(0)=2,y^'(0)=1 ?

The solution for 4y^{''}-2y^'=4x,y(0)=2,y^'(0)=1 is y=10e^{x/2}-8-x^2-4x

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4y^{''}-2y^'=4x,y(0)=2,y^'(0)=1

4 y^{^{''}} - 2 y^{^{'}} = 4 x, y (0) = 2, y^{^{'}} (0) = 1

y = 10 e^{\frac{x}{2}} - 8 - x^{2} - 4 x

Solution steps

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

Solve linear ODE:

y = 10 e^{\frac{x}{2}} - 8 - x^{2} - 4 x

y = 10 e^{\frac{x}{2}} - 8 - x^{2} - 4 x

y^{^{''}} - 4 y^{^{'}} - 12 y = (4 x + 7) e^{- 2 x}

y^{^{''}} + 4 y^{^{'}} + 5 y = 2 e^{3 x}

y^{^{'''}} - 6 y^{^{''}} + 12 y^{^{'}} - 8 y = 6 e^{2 x} + 16 x^{2}

y^{^{''}} + 4 y = \frac{3}{sin ( 2 x )}

a y^{^{''}} + 2 y^{^{'}} = 4 e^{2 t}, y (0) = 0, y^{^{'}} (0) = 0

What is the solution for 4y^{''}-2y^'=4x,y(0)=2,y^'(0)=1 ?
The solution for 4y^{''}-2y^'=4x,y(0)=2,y^'(0)=1 is y=10e^{x/2}-8-x^2-4x