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Popular >

(x+e^x)ydx-dy=0

Solution

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

Solution

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

Solution steps

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

Solve separable ODE:

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

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

Popular Examples

\frac{df}{d x} = f (x)

(4x+xy^2)dx+(y+x^2y)dy=0,y(1)=2

(4 x + x y^{2}) d x + (y + x^{2} y) d y = 0, y (1) = 2

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

(dy)/(dx)=(5(1+x))/(xy)

\frac{d y}{d x} = \frac{5 ( 1 + x )}{x y}

(8x+2)dx+(15y^2+3)dy=0,y(1)=1

(8 x + 2) d x + (15 y^{2} + 3) d y = 0, y (1) = 1

Frequently Asked Questions (FAQ)

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