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Equation: u$_x$ + $\frac{x}{y}$u$_y$ = 0. Initial condition: u(0, y) = exp(-y$^2$)

My professor found the solution u(x, y) = f(y$^2$ - x$^2$) using characteristic curves

He then evaluated u(0, y) = f(y$^2$)

His final answer was u(x, y) = exp(-(y$^2$-x$^2$)).

I understand everything up until the last two lines - how did he get from the $2$nd to last line to the final line of working?

Alessio K
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juliana paternoster
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1 Answers1

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The general solution is $u(x,y)=f(y^2-x^2)$ for any function $f$ (which will be determined by the initial condition). Then using the initial condition we obtain

$$u(0,y)=f(y^2)=e^{-y^2}$$

or $f(v)=e^{-v},$ so we have found $f$ which satisfies the initial condition. Then we have the final solution

$$u(x,y)=f(y^2-x^2)=e^{-(y^2-x^2)}$$

Alessio K
  • 10,599