Let x=(x^ y^) be the solution of 2x^2y,dy-(1-xy^2),dx=0, x0, y(2)=_e2.

Question

Let $\alpha x=\exp(x^\beta y^\gamma)$ be the solution of $2x^2y\,dy-(1-xy^2)\,dx=0$, $x>0$, $y(2)=\sqrt{\log_e2}$. Then $\alpha+\beta-\gamma$ equals:

Accepted Answer

With $y^2=t$: $x^2\dfrac{dt}{dx}=1-xt \implies \dfrac{dt}{dx}+\dfrac{t}{x}=\dfrac{1}{x^2}$. I.F.$=x$.

$$tx = \ln x+c \implies xy^2 = \ln(2x).$$

So $2x = e^{xy^2}$, giving $\alpha=2,\beta=1,\gamma=2$, and $\alpha+\beta-\gamma=1$.

Solutions · 1