# What is a general solution to the differential equation #y'=sqrtxe^(4y)#?

This is a separable differential equation.

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The general solution to the differential equation ( y' = \sqrt{x} e^{4y} ) can be obtained by separating variables and integrating.

Starting with the given differential equation:

( y' = \sqrt{x} e^{4y} )

Separate variables:

( e^{-4y} dy = \sqrt{x} dx )

Integrate both sides:

( \int e^{-4y} dy = \int \sqrt{x} dx )

( -\frac{1}{4} e^{-4y} = \frac{2}{3} x^{\frac{3}{2}} + C )

Multiply both sides by -4:

( e^{-4y} = -\frac{8}{3} x^{\frac{3}{2}} + C )

Take the natural logarithm of both sides to solve for ( y ):

( -4y = \ln\left(-\frac{8}{3} x^{\frac{3}{2}} + C\right) )

( y = -\frac{1}{4} \ln\left(-\frac{8}{3} x^{\frac{3}{2}} + C\right) )

Where ( C ) is the constant of integration. This is the general solution to the given differential equation.

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