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Popular >

x+xy+y^'=0

Solution

x + x y + y^{^{'}} = 0

Solution

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

Solution steps

x + x y + y^{'} (x) = 0

Solve separable ODE:

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

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

Popular Examples

(dy)/(dx)=2e^{x-y},y(0)=ln(4)

\frac{d y}{d x} = 2 e^{x - y}, y (0) = ln (4)

(dy)/(dx)=(4y)/(1+x)

\frac{d y}{d x} = \frac{4 y}{1 + x}

y^'=((2y+3))/(x^2)

y^{^{'}} = \frac{( 2 y + 3 )}{x ^{2}}

(dy)/(dx)=2*(x^2)*(y^2)

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

dx-(dy)/(1+y^2)=0

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

Frequently Asked Questions (FAQ)

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