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Popular >

yy^'=2xy+2x

Solution

y y^{^{'}} = 2 x y + 2 x

Solution

y = - W (- \frac{1}{e ^{- c_{1} + x^{2}}}) - 1

Solution steps

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

Solve Bernoulli ODE:

y = - W (- \frac{1}{e ^{- c_{1} + x^{2}}}) - 1

y = - W (- \frac{1}{e ^{- c_{1} + x^{2}}}) - 1

Popular Examples

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

y^{^{'}} + \frac{- 2 x y}{1 + x ^{2}} = 0

(dy)/(dx)=-4e^{x+y}

\frac{d y}{d x} = - 4 e^{x + y}

y^{2} - x^{2} y^{^{'}} = 0

(dy)/(dx)=(1-y^2)/(1+x^2)

\frac{d y}{d x} = \frac{1 - y ^{2}}{1 + x ^{2}}

y^'+((2x)/(x^2-1))y= x/(-1+x^2)

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

Frequently Asked Questions (FAQ)

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