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How to solve a differential equation of the following form:

$\dfrac{ax}{\dfrac{\partial{f(x,y)}}{\partial x}} + \dfrac{by}{\dfrac{\partial{f(x,y)}}{\partial y}}=-1 $

Where, $f(x_0,y_0)=c$.

2 Answers 2

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$$\dfrac{ax}{\dfrac{\partial{f(x,y)}}{\partial x}} + \dfrac{by}{\dfrac{\partial{f(x,y)}}{\partial y}}=-1 \quad \text{ with condition} \quad f(x_0,y_0)=c$$ This is a non-linear PDE : The given condition isn't sufficient to determine a unique solution.

For example, the method of separation of variables leads to : $$f(x,y)=\frac{a(x^2-x_0^2)}{2\lambda}-\frac{b(y^2-y_0^2)}{2(\lambda+1)}+c$$ which is a family of solutions of the PDE fitting with the given condition.

This is true any constant $\lambda$, except $0$ and $-1$. So, they are an infinity of solutions.

Note : The solution $\frac a2(x^2-x_0^2)-\frac b{\color{red}{4}}(y^2-y_0^2)+c$ cited by "polfosol" is the particular solution in case of $\lambda=1$.

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    Note $\frac{b(y^2-y^2_0)}{2\lambda}-\frac{a(x^2-x^2_0)}{2(\lambda+1)} + C$ are also solutions – 2017-01-08
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    This is exactly the same family of solutions since $\lambda$ is any constant except $0$ and $-1$ : Simply replace $\lambda$ by $(-1-\lambda)$. – 2017-01-08
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    @ JJacquelin Could you please show how you got $\lambda$ using the method of separation. – 2017-01-08
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    They are a choice of manners to separate the variables. One of them is $f(x,y)=F(x)+G(y)$. Putting into the PDE leads to $\frac{ax}{F'(x)}=1-\frac{by}{G'(y)}$ which implies that both terms are constant, namely $=\lambda$. Then, integrate and you obtain the above result. – 2017-01-09
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Using the definition of implicit differentiation for $f(x,y)$ one can write $$\frac{dy}{dx}=-\frac{f_x}{f_y}$$ And from your equation: $$\frac{a x}{f_x}=-\frac{b y+f_y}{f_y}$$ Therefore: $$\begin{align} \frac{dy}{dx}=\frac{a x}{b y+f_y}&\implies\int a x\,dx=\int(b y+f_y)dy\\ &\implies\dfrac 12 ax^2+C=\dfrac 12 by^2+f(x,y)\\ &\implies f(x,y)=\dfrac 12 ax^2-\dfrac 12 by^2+C \end{align}$$ Update:
The boundary condition is $f(x_0,y_0)=c$ which means $$\dfrac 12 (ax_0^2-by_0^2)+C=c\implies C=c-\dfrac 12 (ax_0^2-by_0^2)$$ and finally $$f(x,y)=\dfrac 12 a(x^2-x_0^2)-\dfrac 12 b(y^2-y_0^2)+c$$ But it seems the answer must be $\frac a2(x^2-x_0^2)-\frac b{\color{red}{4}}(y^2-y_0^2)+c$. Don't know why I ended up with that.

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    Hm $\frac{1}{k}ax^2-\frac{1}{k+2}by^2+C$ and $\frac{1}{k}bx^2-\frac{1}{k+2}ay^2+C$ for any $k\neq 0,-2$ are all valid solutions. I'm not sure I understand your method of solving by implicit differentiation formulas. – 2017-01-08
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    @GrantB. I think the mistake was where I put $$\int f_y dy=f(x,y)$$ – 2017-01-08
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    @polfosol, I am interested to see what went wrong in your solution. – 2017-01-10
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    @usr109876787 As I said, the wrong part was the fact that $\int f_y dy\ne f(x,y)$. Try some examples to see what I'm saying – 2017-01-10