Question

Find the particular solution of the differential equation dy/dx = (x-2)e^(-2y) satisfying the initial condition y(2)=ln(2)

Answer 1

Hmm..well it's separable...so divide both sides by \(\large e^{-2y}\)

\[\large e^{2y}\frac{dy}{dx} = (x - 2)\]

Then integrate both sides

Answer 2

but how does that satisfy the initial condition?

Answer 3

I've gotten out -(ln|-x^2+4x|)/2 but where to go from here?

Answer 4

The initial condition is what we put in at the end to solve for the constant...

\[\large \frac{1}{2} \log(x^2 - 4x + C_1)\]

is what I get..remember the +C when integrating

Answer 5

And from there is where you apply the initial condition

y(2) = ln(2)

so

\[\large \ln(2) = \frac{1}{2}\ln((2)^2 - 4(2) + C_1)\]
\[\large \ln(2) = \frac{1}{2}\ln(-4 + C_1)\]
\[\large 2\ln(2) = \ln(C_1 - 4)\]
\[\large 4 = C_1 - 4\]
\[\large C_1 = 8\]

Now just make that your constant...

\[\large y(x) = \frac{1}{2}\log(x^2 - 4x + 8)\] would be your final answer