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Popular >

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

Solution

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

Solution

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

Solution steps

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

Solve separable ODE:

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

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

Graph

Popular Examples

cos^2(x)(dy)/(dx)=y^2sin^2(x)

cos^{2} (x) \frac{d y}{d x} = y^{2} sin^{2} (x)

(-y-2)dx+(x-2)dy=0

(- y - 2) d x + (x - 2) d y = 0

(dx)/(dt)x=t^2-5

\frac{d x}{d t} x = t^{2} - 5

(dy)/(dt)=100-y,y(0)=20

\frac{d y}{d t} = 100 - y, y (0) = 20

(dx)/(dt)=30-11x+x^2

\frac{d x}{d t} = 30 - 11 x + x^{2}

Frequently Asked Questions (FAQ)

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