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Popular >

(x+1)dx+e^ydy=0

Solution

(x + 1) d x + e^{y} d y = 0

Solution

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

Solution steps

(x + 1) d x + e^{y} d y = 0

Solve separable ODE:

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

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

Popular Examples

y^{^{'}} = y - \frac{y ^{2}}{3}

(dy)/(dx)=(1+2y)/(sqrt(4-x^2))

\frac{d y}{d x} = \frac{1 + 2 y}{4 - x ^{2}}

(dy}{dx}=\frac{x^5)/y

\frac{d y}{d x} = \frac{x ^{5}}{y}

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

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

(dy)/(dx)=-x/y ,y(1)=2

\frac{d y}{d x} = - \frac{x}{y}, y (1) = 2

Frequently Asked Questions (FAQ)

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