What is the solution for 2y^{''}-3y^'=9x^2+10,y(0)=1,y^'(0)=-3 ?

The solution for 2y^{''}-3y^'=9x^2+10,y(0)=1,y^'(0)=-3 is y=2e^{(3x)/2}-1-x^3-2x^2-6x

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2y^{''}-3y^'=9x^2+10,y(0)=1,y^'(0)=-3

2 y^{^{''}} - 3 y^{^{'}} = 9 x^{2} + 10, y (0) = 1, y^{^{'}} (0) = - 3

y = 2 e^{\frac{3 x}{2}} - 1 - x^{3} - 2 x^{2} - 6 x

Solution steps

2 y^{''} (x) - 3 y^{'} (x) = 9 x^{2} + 10

Solve linear ODE:

y = 2 e^{\frac{3 x}{2}} - 1 - x^{3} - 2 x^{2} - 6 x

y = 2 e^{\frac{3 x}{2}} - 1 - x^{3} - 2 x^{2} - 6 x

x^{^{''}} + 2 x^{^{'}} + x = 3 cos (t)

y^{^{''}} + 2 y^{^{'}} + 4 y = 5 sin (3 t)

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

y^{^{''}} + 2 y^{^{'}} + 5 y = 1

y^{^{''}} - 4 y^{^{'}} + 4 y = x^{2} e^{2 x}

What is the solution for 2y^{''}-3y^'=9x^2+10,y(0)=1,y^'(0)=-3 ?
The solution for 2y^{''}-3y^'=9x^2+10,y(0)=1,y^'(0)=-3 is y=2e^{(3x)/2}-1-x^3-2x^2-6x