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Popular >

y^'=(2e^x)/(e^y)

Solution

y^{^{'}} = \frac{2 e ^{x}}{e ^{y}}

Solution

y = ln (2 e^{x} + c_{1})

Solution steps

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

Solve separable ODE:

y = ln (2 e^{x} + c_{1})

y = ln (2 e^{x} + c_{1})

Popular Examples

(y+1)cos(x)dx-dy=0

(y + 1) cos (x) d x - d y = 0

derivative of x^4+y^4=4,x^3+4y^3y^'=0

\frac{d}{d x} (x^{4} + y^{4}) = 4, x^{3} + 4 y^{3} y^{^{'}} = 0

3(1+q^2)(dp)/(dq)=2pq(p^3-1)

3 (1 + q^{2}) \frac{d p}{d q} = 2 pq (p^{3} - 1)

((3x^2+1))/(2y-3)=(dy)/(dx)

\frac{( 3 x ^{2} + 1 )}{2 y - 3} = \frac{d y}{d x}

xy^2(dy)/(dx)=x+1

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

Frequently Asked Questions (FAQ)

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