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Solve $u_x+(1+y^2)u_y=0$

I wanted to use characteristic method so I have that $dx/1=dy/(1+y^2)=du/0$

Which gives $dy/dx=1+y^2$ solving gives $-arctan(x)=x+c$

And solving $du/dx=0$ gives $u=g(c)$

Im not sure what I'm supposed to do now.

AColoredReptile
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  • You've missed a negative sign for $(1+y^2)$,and your integration of $\frac{dy}{1+y^2}$is wrong, $$\int \frac{-dy}{1+y^2}=- \tan^{-1}{y}$$ – Smriti Sivakumar Jan 29 '21 at 04:04

1 Answers1

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Since that we need to solve $$\color{blue}{1}\frac{\partial u}{\partial x}(x,y)+\color{blue}{(1+y^{2})}\frac{\partial u}{\partial y}(x,y)=\color{blue}{0}$$ then the Lagrange-Charpit equations are given by $$\frac{dx}{1}=\frac{dy}{1+y^{2}}=\frac{du}{0}$$ Now, we have $$\frac{dx}{1}=\frac{dy}{1+y^{2}} \quad [1] \quad \text{and} \quad \frac{dx}{1}=\frac{du}{0} \quad [2]$$ In $[1]$ we can see that $$\int \frac{dx}{1}=\int \frac{dy}{1+y^{2}} \implies \color{blue}{\boxed{x=\arctan(y)+C_{1}}}$$ and in $[2]$ we can see that $$\frac{du}{dx}=0 \implies \color{blue}{\boxed{u=C_{2}}}$$ The last equation gives inmediately $u=C_{2}$.

The general solution is given by $$C_{2}=f(C_{1})$$ where $f$ is an arbitrary function of the single variable $C_{1}$. Substituting $C_{1}$ and $C_{2}$ by their expressions, we finally have that $$\color{blue}{\boxed{u(x,y)=f(x-\arctan(y))}}$$