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\begin{document}

{\bf Question}

Using Lagrange's method, obtain the general solution of
$uu_x+u_y=1$. If $u=0$ on $y^2=2x$ obtain the solution and state
the region in which it is valid.

\medskip

{\bf Answer}

Lagrange gives

$\ \ \ds\frac{dx}{d\xi}=u,\ \ds\frac{dy}{d\xi}=1,\
\ds\frac{du}{d\xi}=1$

$\Rightarrow u=\xi+a,\ y=\xi+b \ \ \ (1)$

$\Rightarrow \ds\frac{dx}{d\xi}=\xi+a$

$\Rightarrow x=\ds\frac{1}{2}\xi^2+a\xi+c\ \ \ (2)$

where $a,\ b,\ c$ are constants

Therefore we have ((1)): $u-y=const\ \ \ (3)$

and ((2)): $x=\ds\frac{1}{2}(u-a)^2+a(u-a)-+c$

$\Rightarrow x=\ds\frac{1}{2}u^2+const$

$\Rightarrow \left(x-\ds\frac{1}{2}u^2\right)=const\ \ \ (4)$

Therefore (3) and (4) give general solution

$f\left(u-y,\ x-\ds\frac{1}{2}u^2\right)=const$

Write this as $u-y=g\left(x-\ds\frac{1}{2}u^2\right)$ say.

Boundary condition then gives $-\sqrt{2x}=g(x)$, so

$g\left(x-\ds\frac{1}{2}u^2\right)=-\sqrt{2x-u^2}$.

Thus $u-y=-\sqrt{2x-u^2} \longrightarrow u=\ds\frac{y}{2} \pm
\left(x-\ds\frac{y^2}{4}\right)^{\frac{1}{2}}$.

Since $u=0$ on $y^2=2x$ we need negative root, so

$$\un{u=\ds\frac{1}{2}y-\left(x-\ds\frac{u^2}{4}\right)}$$

No solution for $x<\ds\frac{y^2}{4}$.

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