# Math Help - First-order linear ode

1. If you change the K in the denominator to +K, then the answer is y = -2...
No, because the k in the numerator and the k in the denominator are the same k. They would both have to change.

I agree with you up to here:

$\displaystyle -2\,\frac{e^{-\frac{1}{2}x^2} + K}{e^{-\frac{1}{2}x^2} - K}.$

Replacing $K$ with $-K$ yields

$\displaystyle -2\,\frac{e^{-\frac{1}{2}x^2} - K}{e^{-\frac{1}{2}x^2} + K}.$

$\displaystyle 2\,\frac{K-e^{-\frac{1}{2}x^2}}{K+e^{-\frac{1}{2}x^2}}.$
Now, finally, one more trick: replacing $K$ with $1/K,$ will, I think, finish off the problem.
3. Well, you could take the previous step, right before replacing $K$ with $-K$, and then let $K=0$ there. Equivalently, let $K\to\infty$ in the final solution.