### Author Topic: Assignment 2.2 Problem3 (16)  (Read 961 times)

#### Wanying Zhang

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##### Assignment 2.2 Problem3 (16)
« on: January 24, 2019, 12:25:14 AM »
Given the question $u_t + 3u_x -2u_y = xyu$. My steps are:
$$\frac{dt}{1} = \frac{dx}{3} = \frac{dy}{-2} = \frac{du}{xyu} \Rightarrow x=C_1+3t, y=C_2-2t$$
$$xydt = \frac{du}{u} \Rightarrow (C_1+3t)(C_2-2t)dt = \frac{du}{u}$$
$$\ln u = C_1C_2 t + \frac{t^2}{2} (3C_2 - 2C_1) - 2t^3 +\phi(C_1, C_2)$$
I know then I need to insert base $e$ into both sides to get $u$, but it would be complex. I wonder is there any simple way to solve this? Or do I have something wrong in above equations so that leads to the complex solution?

#### Victor Ivrii

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##### Re: Assignment 2.2 Problem3 (16)
« Reply #1 on: January 24, 2019, 06:16:36 AM »
What you really need to do is to replace $C_1,C_2$ by their expression through $x,y,t$

#### Jerry Qinghui Yu

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##### Re: Assignment 2.2 Problem3 (16)
« Reply #2 on: January 24, 2019, 10:27:46 AM »
Could you just post the solution? Could really learn from examples instead of words.

#### Victor Ivrii

Given the question $u_t + 3u_x -2u_y = xyu$. My steps are:
$$\frac{dt}{1} = \frac{dx}{3} = \frac{dy}{-2} = \frac{du}{xyu} \Rightarrow x=C_1+3t, y=C_2-2t$$
$$xydt = \frac{du}{u} \Rightarrow (C_1+3t)(C_2-2t)dt = \frac{du}{u}$$
$$\ln u = C_1C_2 t + \frac{t^2}{2} (3C_2 - 2C_1) - 2t^3 +\phi(C_1, C_2)$$
Since $C_1=x-3t$, $C_2=y+2t$ we plug
$$u= f(x-3t, y+2t) \exp \bigl( (x-3t)(y+2t)t +\frac{t^2}{2}(3y+6t-2x+6t) -2t^3\bigr)$$
with an arbitrary function $f.,.)$ of two variables ($f=e^\phi$).