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Popular >

(x^2+4)y^'=xy

Solution

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

Solution

y = c_{1} x^{2} + 4

Solution steps

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

Solve separable ODE:

y = c_{1} x^{2} + 4

y = c_{1} x^{2} + 4

Graph

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limit as x approaches 1 of (x^2-x)/(x-1)

x \to 1 lim (\frac{x ^{2} - x}{x - 1})

Frequently Asked Questions (FAQ)

What is the solution for (x^2+4)y^'=xy ?
The solution for (x^2+4)y^'=xy is y=c_{1}sqrt(x^2+4)