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**Home Assignment 2 / Re: Assignment 2.2 Problem3 (16)**

« **on:**January 24, 2019, 11:43:29 AM »

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Given the question $u_t + 3u_x -2u_y = xyu$. My steps are:Since $C_1=x-3t$, $C_2=y+2t$ we plug

$$\frac{dt}{1} = \frac{dx}{3} = \frac{dy}{-2} = \frac{du}{xyu} \Rightarrow x=C_1+3t, y=C_2-2t$$

$$xydt = \frac{du}{u} \Rightarrow (C_1+3t)(C_2-2t)dt = \frac{du}{u}$$

$$\ln u = C_1C_2 t + \frac{t^2}{2} (3C_2 - 2C_1) - 2t^3 +\phi(C_1, C_2)$$

$$

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 ($f=e^\phi$).