Recent content by Mathwebster

What I'm saying is that after your integration, your simplifaction is incorrect. $$\dfrac{e^{\int\frac{1-x}{x}dx}}{x} = \dfrac{e^{ln x - x}}{x} = \dfrac{x e^{-x}}{x} = e^{-x}$$ (I've also be a little relaxed with absolutes).

You might want to check what you did after you integrated.

Make the transformation $$x = - \dfrac{r s'}{s},\;\;\; y = r$$ where $$s = s(r)$$ and see where that takes you.

This would be wonderful if that was true b/c the equation is homogenous. With the right side numerator of power one, the problem is a little more difficult!

If you make the change of variables $$x = r + 1/2,\;\;\; y = e^s$$ where $$s = s(r)$$, your ODE becomes $$s'' = 2 r s'^2$$ which now becomes separable. This should help you.

Sorry, that was a typo. If $$u = h(6x+y) + g(6x-y)$$ then $$u_x = 6h'(6x+y) + 6g'(6x-y)$$ $$u_y = h'(6x+y) - g'(6x-y)$$ So from the original set of PDEs we have $$6h'(6x+y) + 6g'(6x - y) + 4 v_y = 0$$ $$v_x + 9\left(h'(6x+y) - g'(6x-y)\right) = 0$$ or $$v_x = - 9h'(6x+y) + 9 g'(6x-y)$$...

Choose one of them, say $$u = g(x+6y) + g(y-6x)$$, then sub this back into your system. Then integrate each giving the solution for $$v(x,y)$$. Then use your BC's.

Why are you assuming the solutions $u$ and $v$ are exactly the same? As you're seeing this is inconsistent with the initial conditions.

First. I think you should be consistent with your notation. Either use $$x$$ and $$y$$ or $$x_1$$ and $$x_2$$ but not both. It's confusing. Second, as you have the boundary of an ellipse, have you thought of introducing new coordinates $$x = a r \cos \theta, y = b r \sin \theta?$$

Divide your equation by $x$ and then differentiate. You should find that the new equation factors into two pieces that integrate easily. With these, go back to the original equation and check that they both work (and adjust constants accordingly).

Re: Evaluate 2012+((a-b)(b-c)(c-a))/abc My solution

My solution

First, is it $\dfrac{b}{2a}$ or $- \dfrac{b}{2a}$. It really makes a difference. Second, you can use the hyperbolic sinh-cosh identity. Try that.

Yes, that was a typo. It should be $b/2$. As for finding the constant. Go back to the original equation after the substitution, namely $ y'^2 = \mu \sigma^2 y^2 - \dfrac{2 \alpha \sigma^2}{\sigma+2} y - \dfrac{ \gamma \sigma^2}{\sigma+1}$ and substitute your solution.

You can do what you did but try differentiating. In doing so we obtain $2 u' u'' = (2 a u +b) u' $ giving $u ' = 0$ or $u'' = au + b/a$. I think the latter is easier to solve than your approach.