Using method of characteristics to solve pde $xyu_x+u_y+u=4$

Asked Jun 10 '21 at 05:51

Active Jun 10 '21 at 05:51

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My attempt:

$\frac{dx}{xy}=\frac{dy}{1}=\frac{du}{u-4}$

Now take $\frac{dx}{xy}=\frac{dy}{1}\implies \frac{dx}{x}=ydy \implies c_1=\frac{^{e^{\frac{y^2}{2}}}}{x}$

now $\frac{dy}{1}=\frac{du}{u-4}\implies y=\log|u-4|+\log c_2 \implies c_2= e^y(u-4)$

From both i am unable to find $u$

thanks your time..

asked Jun 10 '21 at 05:51

learner

What are the initial conditions? – Stuck Jun 10 '21 at 05:57
@Stuck..no initial conditions – learner Jun 10 '21 at 06:11
Oh.. it seems to me you can just solve for $u$ via $c_2=e^y(u-4),$ no? – Stuck Jun 10 '21 at 06:37
@Stuck.. I think we to find $u$ from $c_1,c_2$ – Inverse Problem Jun 10 '21 at 06:46
1

You made some mistakes. It should be $$\frac{dx}{xy} = \frac{dy}{1} = \frac{du}{\color{green}{4 - u}}$$ and also $$\frac{dx}{x} = y dy \implies \ln x = \frac{y^{2}}{2} + C_{1} \implies x e^{\color{red}{-} y^{2}/2} = C_{1}$$ The other characteristic is $C_{2} = e^{y}(4 - u)$. The general solution is then $$C_{2} = f(C_{1}) \implies e^{y}(4 - u) = f(x e^{- y^{2}/2}) \implies u = 4 - e^{- y} f(x e^{- y^{2}/2})$$ for some differentiable function $f$. – Matthew Cassell Jun 10 '21 at 06:47
@mattos Why is the solution given by $C_2=f(C_1)$? – Stuck Jun 10 '21 at 07:01
@Stuck..Its called implict equation – Inverse Problem Jun 10 '21 at 07:03

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