How do you solve for the equation #dy/dx=(3x^2)/(e^2y)# that satisfies the initial condition #f(0)=1/2#?
La risposta รจ: #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#.
Ora per trovare #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)#.