Solve the Cauchy problem

Asked Jan 29 '14 at 12:20

Active Jan 29 '14 at 13:32

Viewed 230 times

Let $U(x,y)$ be the solution to the Cauchy problem:

$$ xU_x + U_y = 1, U(x,0) = 2\ln(x) , x>1$$

Then $U(e,1) =$

My work is as follows. Since,

$$\frac{dx}{x} = \frac{dy}{1} = \frac{dU}{1},$$

we get $U - y = C_1$ and $U = \ln\frac{x}{C_2}$.

We have given that $U(x,0) = 2 \ln(x)$, so $C_1 = 2\ln(x)$.

Thus, $U(x,y) = y + 2 \ln(x)$ But my answer is $2$, where am I wrong?

edited Jan 29 '14 at 13:32

Martin Sleziak

asked Jan 29 '14 at 12:20

user120386

1

$U(x,y)=-y+2 \ln x$. – user1337 Jan 29 '14 at 12:25
why U(x,y) = -y +2 ln(x) – user120386 Jan 29 '14 at 16:34
Following your method I get $U-y=f(U-\ln x)$, and using the initial condition it is easy to determine $f$. – user1337 Jan 29 '14 at 17:30
using the initial condition we get 2 $ln x$ = f(1), how to conclude that U(x,y) = $-y$ + 2$ln x$ – user120386 Jan 29 '14 at 18:33
Actually we get $2 \ln x-0=f(2 \ln x- \ln x)$. – user1337 Jan 29 '14 at 18:37
Thank you for giving me your valuable time – user120386 Jan 30 '14 at 01:19

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