What is the solution for 2y^{''}+y^'-y=3x+4,y(0)=1,y^'(0)=1 ?

The solution for 2y^{''}+y^'-y=3x+4,y(0)=1,y^'(0)=1 is y=-3x+8e^{x/2}-7

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2y^{''}+y^'-y=3x+4,y(0)=1,y^'(0)=1

2 y^{^{''}} + y^{^{'}} - y = 3 x + 4, y (0) = 1, y^{^{'}} (0) = 1

y = - 3 x + 8 e^{\frac{x}{2}} - 7

Solution steps

2 y^{''} (x) + y^{'} (x) - y = 3 x + 4

Solve linear ODE:

y = - 3 x + 8 e^{\frac{x}{2}} - 7

y = - 3 x + 8 e^{\frac{x}{2}} - 7

y^{^{''}} + 9 y = 18 sin (3 t)

y^{^{''}} - y^{^{'}} - 6 y = 3 e^{2 x}

x^{^{''}} + 25 x = cos (t)

y^{^{''}} + 7 y^{^{'}} - 12 y = e^{3} x - 2 e^{x} - 5 x

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

What is the solution for 2y^{''}+y^'-y=3x+4,y(0)=1,y^'(0)=1 ?
The solution for 2y^{''}+y^'-y=3x+4,y(0)=1,y^'(0)=1 is y=-3x+8e^{x/2}-7