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

Question

Charles Anctil · Accepted Answer

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)#.

Aurora Alvarado · Answer

undefined To solve the differential equation $ \frac{dy}{dx} = \frac{3x^2}{e^{2y}} $ with the initial condition $ f(0) = \frac{1}{2} $, we can separate variables and integrate both sides.

1. Separate variables:
$$ e^{2y} \, dy = 3x^2 \, dx $$

2. Integrate both sides:
$$ \int e^{2y} \, dy = \int 3x^2 \, dx $$

3. Solve the integrals:
$$ \frac{1}{2} e^{2y} = x^3 + C $$
Where $ C $ is the constant of integration.

4. Solve for $ y $:
$$ e^{2y} = 2x^3 + C $$
$$ 2y = \ln(2x^3 + C) $$
$$ y = \frac{1}{2} \ln(2x^3 + C) $$

5. Apply the initial condition $ f(0) = \frac{1}{2} $:
$$ \frac{1}{2} = \frac{1}{2} \ln(C) $$
$$ 1 = \ln(C) $$
$$ C = e $$

6. Substitute $ C = e $ into the solution:
$$ y = \frac{1}{2} \ln(2x^3 + e) $$