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I have not solved PDEs in eons.

I am battling to solve the following:

$ x^2 \frac{\partial s}{\partial x} +xy\frac{\partial s}{\partial y} =1$

I have managed to solve a similar PDE which equals zero, but I don't know how to deal with the 1 in this equation.

If the steps are similar then I would divide by $x^2$ to get $\frac{\partial s}{\partial x} +\frac{y}{x}\frac{\partial s}{\partial y} =\frac{1}{x^2}$

However I don't know where to go from here :(

NOTE: This equation is actually part of a pair of equations that I need to solve in order to determine the canonical coordinates $r(x,y)$ and $s(x,y)$. I'm trying to solve y''=0 using symmetry. I've solved $r(x,y)=\frac{y}{x}$ and now need to get $s(x,y)$.

sarah jamal
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    If you divide by $x$, the left-hand side becomes $x s_x + y s_y$, which is $r s_r$ in polar coordinates. So you could try changing to polar coordinates. – Hans Lundmark Jan 05 '18 at 05:20
  • Solution of homogeneous equation $x^2 \frac{\partial s}{\partial x} +xy\frac{\partial s}{\partial y} =0$ is $s_c=F(y/x)$. Particular solution is $s_p=-1/x$. Then $s=s_c+s_p=F(y/x)-1/x$. – Aleksas Domarkas Jun 20 '18 at 04:28

3 Answers3

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I will follow your approach of dividing through by $x^{2}$. Using the method of characteristics, we obtain the ODEs

\begin{align} \frac{dy}{dx} &= \frac{y}{x} \quad (1) \\ \frac{ds}{dx} &= \frac{1}{x^{2}} \quad (2) \\ \end{align}

Solving $(1)$ yields

\begin{align} y &= y_{0} x \implies y_{0} = \frac{y}{x} \\ \end{align}

Solving $(2)$ yields

\begin{align} s &= \frac{-1}{x} + f(y_{0}) \\ &= \frac{-1}{x} + f \left( \frac{y}{x} \right) \end{align}

where $f$ is an arbitrary differentiable function. You can check via differentiation that this is the general solution.

I would also suggest trying to use Hans' idea and change your problem to polar coordinates.

Matthew Cassell
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$$ x^2 \frac{\partial s}{\partial x} +xy\frac{\partial s}{\partial y} =1$$ System of characteristic ODEs : $\quad\frac{dx}{x^2}=\frac{dy}{xy}=\frac{ds}{1}$

First set of characteristics, from $\quad\frac{dx}{x^2}=\frac{dy}{xy} \quad\implies\quad \frac{y}{x}=c_1$

Second set of characteristics, from $\quad\frac{dx}{x^2}=\frac{ds}{1} \quad\implies\quad s+\frac{1}{x}=c_2$

General solution on implicit form : $\quad F\left(\frac{y}{x} \:,\:s+\frac{1}{x} \right)=0\quad$ where $F$ is any differentiable function of two variables.

Or equivalently, solution on explicit form : $\quad s+\frac{1}{x}=f\left(\frac{y}{x} \right)\quad$ where $f$ is any differentiable function. $$s=-\frac{1}{x}+f\left(\frac{y}{x} \right)$$

JJacquelin
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We start by rewriting the equation,

$$ x \frac{\partial s}{\partial x } + y\frac{\partial s }{\partial y} = \frac{1}{x}$$

in polar coordinates this becomes,

$$ r \cos\theta \frac{\partial s}{\partial x } + r\sin\theta \frac{\partial s }{\partial y} = \frac{1}{r\cos\theta}$$

Note that,

$$ \frac{\partial}{\partial r } = \cos\theta \frac{\partial}{\partial x} + \sin\theta \frac{\partial}{\partial y} ,$$

which allows us to rewrite our equation as,

$$ r\frac{\partial s}{\partial r} = \frac{1}{r\cos\theta}$$

$$ \frac{\partial s}{\partial r} = \frac{1}{r^2\cos\theta}$$

integrating with respect to $r$ we get,

$$ s = -\frac{1}{r\cos\theta} + f(\theta) $$

$$ s = -\frac{1}{x} + f(\theta) $$

Where $f$ is an arbitrary function of $\theta$. We can write replace this with an arbitrary function of $y/x$; this is because a function of $\theta$ can also be written as a function of $\tan\theta = y/x$. We will call this new function $F$, where $F(y/x)=f(\theta)$.

$$ s = -\frac{1}{x} + F(y/x) $$

Spencer
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