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

The solution for (x^2+1)y^{''}-2xy^'+2y=0 is y=e^{c_{1}}x^2-e^{c_{1}}+c_{2}x

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(x^2+1)y^{''}-2xy^'+2y=0

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

y = e^{c_{1}} x^{2} - e^{c_{1}} + c_{2} x

Solution steps

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

Solve using substitution:

y = e^{c_{1}} x^{2} - e^{c_{1}} + c_{2} x

y = e^{c_{1}} x^{2} - e^{c_{1}} + c_{2} x

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\frac{d}{d x} (cos^{4} (2 x^{3} - 5))

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y = x (x + 3) (x - 4), x = 0

What is the solution for (x^2+1)y^{''}-2xy^'+2y=0 ?
The solution for (x^2+1)y^{''}-2xy^'+2y=0 is y=e^{c_{1}}x^2-e^{c_{1}}+c_{2}x