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Popular >

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

Solution

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

Solution

y = \frac{ln ( 2 x ^{2} + c _{1} )}{2}

Solution steps

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

Solve separable ODE:

y = \frac{ln ( 2 x ^{2} + c _{1} )}{2}

y = \frac{ln ( 2 x ^{2} + c _{1} )}{2}

Popular Examples

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

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

(dy)/y =(dx)/((x-9)^2-5^2)

\frac{d y}{y} = \frac{d x}{( x - 9 ) ^{2} - 5 ^{2}}

y^'=-(0.1y)/(0.5-0.025x)

y^{^{'}} = - \frac{0.1 y}{0.5 - 0.025 x}

4 y^{11} + y^{^{'}} = 0

(dp)/(dt)=t^2p-7p+t^2-7

\frac{d p}{d t} = t^{2} p - 7 p + t^{2} - 7

Frequently Asked Questions (FAQ)

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