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

The solution for y^{''}-y^'-2y=4x^2,y(0)=-1,y^'(0)=1 is y=(-5/3+2)e^{2x}+5/3 e^{-x}-2x^2+2x-3

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

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

y = (- \frac{5}{3} + 2) e^{2 x} + \frac{5}{3} e^{- x} - 2 x^{2} + 2 x - 3

Solution steps

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

Solve linear ODE:

y = (- \frac{5}{3} + 2) e^{2 x} + \frac{5}{3} e^{- x} - 2 x^{2} + 2 x - 3

y = (- \frac{5}{3} + 2) e^{2 x} + \frac{5}{3} e^{- x} - 2 x^{2} + 2 x - 3

x \to 16 lim (4 x)

\int_{0}^{5} π (\frac{20 - 4 x}{5})^{2} d x

\int 3 u d u

(8.1) (24.9)

x^{2} y^{^{'}} = y^{2} + x y

What is the solution for y^{''}-y^'-2y=4x^2,y(0)=-1,y^'(0)=1 ?
The solution for y^{''}-y^'-2y=4x^2,y(0)=-1,y^'(0)=1 is y=(-5/3+2)e^{2x}+5/3 e^{-x}-2x^2+2x-3