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Popular >

(x^2+1)y^'=xy

Solution

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

Solution

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

Solution steps

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

Solve separable ODE:

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

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

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\int x^{3} - 6 x^{2} + 8 x d x

limit as x approaches 0 of e^{3x}-1

x \to 0 lim (e^{3 x} - 1)

integral of 4xe^{-2x}

\int 4 x e^{- 2 x} d x

Frequently Asked Questions (FAQ)

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