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$u_x+x^2y^4u_y=0 , u(1,y)=cos(2y) \\ \frac{dy}{dx}= x^2y^4 \implies \int y^{-4} dy = \int x^2 dx \implies -\frac{y^{-3}}{3}= \frac{x^3}{3} +C \\ C= -1/3(x^3+y^{-3}) \\ u(x,y)=f(C)=f(-1/3(x^3+y^{-3})) \\ \text{Given auxiliary condition: } u(1,y)= f(\frac{-1}{3}(1+y^{-3}))= \cos(2y).$

Solving for the general form of $f$ is where I get stuck.

I am unsure if I can simplify this problem by using the fact that $C$ is a constant. In other words can is the following a valid step:

$C= -1/3(x^3+y^{-3}) \implies C=(x^3+y^{-3})$ Doing so would mean I would then proceed with $f(1+y^{-3})= \cos(2y)$ and thus solve this form of $f$.

El Spiffy
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2 Answers2

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You just need to say that:

$$ u(x,y)=f\big(x^{3}+y^{-3}\big)$$ and

because your solution to $\ \displaystyle{\frac{dx}{dt}=1}$, $\displaystyle{\frac{dy}{dt}=x^{2}y^{4}}$ and $\displaystyle{\frac{du}{dt}=1}$

is

So, put $x=1$ in $\ u(x,y)=f\big(x^{3}+y^{-3}\big)$ and you will have:

$$ \cos(2y)=u(1,y)=f(1+y^{-3}) $$

then you call $w=1+y^{-3}$ and you can isolate $y$ in terms of $w$ to get:

$$f(w)=\cos\left(2\sqrt[3]{\frac{1}{w-1}}\right)$$

so, your final solution should be:

$$ u(x,y)=\cos\left(2\sqrt[3]{\frac{1}{x^{3}+y^{-3}-1}}\right). $$

Hector Blandin
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  • Thank you for that clarification. As a follow up question, what is that process called where you isolated $y$ and put it in terms of $w$? I seem to be having trouble doing that step and would like to practice it. – El Spiffy Nov 02 '17 at 02:38
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    I would call that process using initial condition to find the function $f$. Recall that $f$ is an arbitrary function of one variable composed with a 2 variable function $x^{3}+y^{-3}$ when you don't have a given initial condition otherwise the formula for that $f$ depends on what is $f(1,y)$ in terms of $y$. This is the part where you use initial conditions or boundary condition depending in what problem you are facing. – Hector Blandin Nov 02 '17 at 02:42
  • I understand that, I guess the manipulation of the variables is what I struggled with. Like for example: $f(2x)= sin(x)$. Setting $w=2x \implies x=1/2w $ and what then? – El Spiffy Nov 02 '17 at 02:44
  • Your welcome. Probably this file can help you http://ramanujan.math.trinity.edu/rdaileda/teach/s14/m3357/lectures/lecture_1_21_slides.pdf – Hector Blandin Nov 02 '17 at 02:46
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$$u_x+x^2y^4u_y=0 \tag 1$$ With the method of characteristics : $$\frac{dx}{1}=\frac{dy}{x^2y^4}=\frac{du}{0}$$ A first family of characteristic curves comes from necesserarily $du=0 \quad\implies\quad u=c_1$

A second family of characteristic curves comes from $\frac{dx}{1}=\frac{dy}{x^2y^4}$ which is a separable ODE easy to solve : $x^3+\frac{1}{y^3}=c_2$

The general solution of Eq.1 expressed on implicit form is : $$\Phi\left(\left(x^3+\frac{1}{y^3}\right)\;,\:u\right)=0$$ where $\Phi(X,Y)$ is any differentiable function of two variables.

Or, equivalent, on explicit form : $$u=F\left(x^3+\frac{1}{y^3}\right) \tag 2$$ where $F(X)$ is any differentiable function.

Then we have to determine which function $F(X)$ satisfies the condition $u(1,y)=\cos(2y)$ $$F\left(1+\frac{1}{y^3}\right)=\cos(2y)$$ In particular $X=1+\frac{1}{y^3}\quad\to\quad y=(X-1)^{-1/3}\quad\to\quad F(X)= \cos(2(X-1)^{-1/3})$

So F(X) is determined. Putting it into Eq.(2) where $X=x^3+\frac{1}{y^3}$ leads to : $$u=\cos\left(2\left(x^3+\frac{1}{y^3}-1\right)^{-1/3}\right)$$

JJacquelin
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