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

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

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

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

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

Solution steps

(1 + x^{2}) 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

\frac{d ^{3}}{d x ^{3}} (ln (x + 2))

f (t) = 9 t^{2} - t

\int \frac{8 x ^{2}}{9 - x ^{2}} d x

- 2 x^{2} + 3, 0 x

\frac{t ^{2}}{t - 11}

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