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Popular >

xy^'e^x+2ye^x=0

Solution

x y^{^{'}} e^{x} + 2 y e^{x} = 0

Solution

y = \frac{c _{1}}{x ^{2}}

Solution steps

x y^{'} (x) e^{x} + 2 y e^{x} = 0

Solve separable ODE:

y = \frac{c _{1}}{x ^{2}}

y = \frac{c _{1}}{x ^{2}}

Graph

Popular Examples

(\partial)/(\partial x)(ln(x^3+y))

\frac{\partial}{\partial x} (ln (x^{3} + y))

tangent of-3x^2,\at x=2 tangent of

- 3 x^{2}, at x = 2

integral of (3x+5)^2

\int (3 x + 5)^{2} d x

derivative of x^2-3x-5x^{-1}+7x^{-2}

\frac{d}{d x} (x^{2} - 3 x - 5 x^{- 1} + 7 x^{- 2})

(dy)/(dx)=(x^2-3)/(6y^2)

\frac{d y}{d x} = \frac{x ^{2} - 3}{6 y ^{2}}

Frequently Asked Questions (FAQ)

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