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

The solution for (1/4)y^{''}+y^'+y=x^2-2x is y=c_{1}e^{-2x}+c_{2}xe^{-2x}+x^2-4x+7/2

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(1/4)y^{''}+y^'+y=x^2-2x

(\frac{1}{4}) y^{^{''}} + y^{^{'}} + y = x^{2} - 2 x

y = c_{1} e^{- 2 x} + c_{2} x e^{- 2 x} + x^{2} - 4 x + \frac{7}{2}

Solution steps

(\frac{1}{4}) y^{''} (x) + y^{'} (x) + y = x^{2} - 2 x

Solve linear ODE:

y = c_{1} e^{- 2 x} + c_{2} x e^{- 2 x} + x^{2} - 4 x + \frac{7}{2}

y = c_{1} e^{- 2 x} + c_{2} x e^{- 2 x} + x^{2} - 4 x + \frac{7}{2}

\frac{d x}{1 - x ^{2}} = \frac{d y}{1 - y ^{2}}

\int \frac{7 x}{( x ^{2} + 7 ) ^{2}} d x

\int (x^{3} + 5)^{2} d x

(x + 4)^{6} (x - 3)^{3}

cos (tan^{3} (x))

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