Separable equations

Question

Find the solution of the differential equation that satisfies the given initial condition.

dy/dx =  5xe^y,    y(0) = 0 .

Answer 1

dy/dx = 5xe^y

e^-ydy = 5xdx

∫ e^-ydy = ∫ 5xdx

-e^-y = (5/2)x² + C

Since (0,0) is on the curve, -e⁰ = 0 + C. So, -1 = 0 + C. Therefore, C = -1.

-e^-y = (5/2)x² - 1

e^-y = (-5/2)x² + 1

-y = ln[(-5/2)x² + 1]

y = -ln[(-5/2)x² + 1]

Answer 2

So for separable equations, we want to "separate" the dy and the dx from each other and get them on their own sides for integrating against. So with dy/dx = 5x e^y, we can divide by e^y and multiply by dx and have 1/e^y dy = 5x dx. Remember that 1/e^y = e^-y by exponent laws. So then we just need to take integrals:

∫ e^-y dy = ∫ 5x dx

-e^-y = 5/2 x²+ C

e^-y = -5/2 x² + C' (where C' = -C. It's totally fine to just keep + C instead of creating a new constant, as it's understood what you mean since C is currently just some arbitrary constant)

-y = ln (-5/2 x² + C')

y = -ln(-5/2 x² + C')

Now we know that y(0) = 0. Therefore 0 = -ln(-5/2 0² + C'), so 0 = -ln(C') = ln(C') if we divide by 0. ln(1) = 0, so C' = 1. Thus we have the solution as:

y = -ln(5/2 x² + 1)