Hi All,
I have 2 problems that I am stuck on (see attached), I dont know how to go any further. Any tips or advice on what Im doing wrong?
Thanks
You're not doing anything wrong - just not going far enough.
(1) You have
$\displaystyle \dfrac{du}{dx} = \dfrac{-2xu}{x^2 + 3B^2u^2+3u^3}$
If you write it as
$\displaystyle \dfrac{dx}{du} = \dfrac{x^2 + 3B^2u^2+3u^3}{-2xu}$. It's Bernoulli in $\displaystyle x$.
(2) You have
$\displaystyle \dfrac{d(x+y)}{y-x} = \dfrac{du}{x-y}$ which integrates giving $\displaystyle x+y+u = c_1$.
Then
$\displaystyle \dfrac{d(x-y)}{x+ y - 2u} = \dfrac{du}{x-y}$
or
$\displaystyle (x-y)d(x-y) = (c_1 - 3u)du $.
This also integrates.
Multiplying by $\displaystyle 2x$ gives
$\displaystyle \dfrac{2x dx}{du} = \dfrac{x^2 + 3B^2u^2+3u^3}{-u}$
and letting $\displaystyle X = x^2$ gives
$\displaystyle \dfrac{dX}{du} = \dfrac{X + 3B^2u^2+3u^3}{-u}$
or
$\displaystyle \dfrac{dX}{du} + \dfrac{X}{u} = - 3B^2u-3u^2$. This is now linear!
First replace your $\displaystyle B$ with $\displaystyle B = \dfrac{y}{u}$ so your solution is
$\displaystyle \dfrac{u}{y} = f\left(x^2u+y^2u + \frac{3}{4}u^4\right)$
or
$\displaystyle x^2u+y^2u + \frac{3}{4}u^4\right = f\left(\dfrac{u}{y}\right)$.
Are you sure about the IC b/c with $\displaystyle u(x,2x) = 0$, the only solution is $\displaystyle u \equiv 0$.
You are right Danny...once again! The IC was for another similar question and this question is to find general solution only.
However shouldnt the GS be
$\displaystyle \dfrac{y}{u} = f\left(x^2u+y^2u + \frac{3}{4}u^4\right)$
or
$\displaystyle x^2u+y^2u + \frac{3}{4}u^4\right = f\left(\dfrac{y}{u}\right)$
ie, you had u/y
Assuming I am correct, how do you know that the only soltion is $\displaystyle u \equiv 0$
As for your first question, it doesn't matter. Define
$\displaystyle f(\lambda) = \bar{f}\left(\dfrac{1}{\lambda}\right)$ then drop the bar.
For your second question, let go back when we introduced new coordiantes such that
$\displaystyle u_s = u_x x_s + u_y y_s$ gave your PDE by choosing
$\displaystyle x_s=x^2 + 3y^2 + 3u^3$
$\displaystyle y_s =-2xy$
$\displaystyle u_s = -2xu$
with the IC as $\displaystyle x = r, y = 2r, u = 0$ when $\displaystyle s = 0.$ From the second two we find
$\displaystyle y u_s - u y_s = 0$ which integrates gives $\displaystyle \dfrac{u}{y} = a(r)$. Imposing the IC gives $\displaystyle a(r) = 0$ giving $\displaystyle u = 0$.