How do you solve for the equation #dy/dx=(3x^2)/(e^2y)# that satisfies the initial condition #f(0)=1/2#?

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Answer 1 · 2015-02-22T17:49:43

The answer is: #y=1/2ln(2x^3+e)#.

First of all, I think ther is a mistake in your writing, I think you wanted to write:

#(dy)/(dx)=(3x^2)/e^(2y)#.

This is a separable differential equations, so:

#e^(2y)dy=3x^2dxrArrinte^(2y)dy=int3x^2dxrArr#

#1/2e^(2y)=x^3+c#.

Now to find #c# let's use the condition: #f(0)=1/2#

#1/2e^(2*1/2)=0^3+crArrc=1/2e#.

So the solution is:

#1/2e^(2y)=x^3+1/2erArre^(2y)=2x^3+erArr2y=ln(2x^3+e)rArr#

#y=1/2ln(2x^3+e)#.