### Recent Posts

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##### Home Assignment 2 / Re: problem4 (20)
« Last post by Brittany Palandra on January 19, 2019, 07:24:22 PM »
We can make the substitution $y = 3x - C$ because we are restricting $u(x, y)$ to the characteristic curves, so I believe we can treat $y$ as equal to $3x - C$ when finding the general solution. We do this because we need the $xydx$ totally in terms of $x$ or we will not be able to integrate both sides. After integrating, we have to get rid of $C$ by replacing it with $3x-y$ again because we want our final solution $u$ to be a function of $x$ and $y$, not of $C$. $C$ is just a constant but it is still in terms of $x, y$ by the characteristic curves.

$C$ is a constant only along integral curves. V.I.
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##### Home Assignment 2 / Re: problem4 (20)
« Last post by Victor Ivrii on January 19, 2019, 07:03:34 AM »
Please learn how to post math properly  Also, asking for help, copy the problem.
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##### Home Assignment 2 / problem4 (20)
« Last post by Zhiman Tang on January 18, 2019, 11:53:36 PM »
the expression on the right hand contains both x and y,
the characteristic line is 3x-y=c
then, we evaluate du = xydx. in this step, we have to replace y by 3x-c before integrating both sides.
My question is, why we have to do so?
And after integrte both sides, why we have to replace c back by 3x-y?
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##### Home Assignment 2 / Re: Home Assignment 2 Problem 2(a)
« Last post by Victor Ivrii on January 18, 2019, 09:31:56 PM »
Since integral curves are rays (straight half-lines) from $(0,0)$  the solution in the plane wit the punched out origin is any function, constant along these rays, in particular $u=f(y/x)$.

But if we want solution in the whole plane, $u$ must be continuous at $(0,0)$ and since all rays intersect there $u$ is just a constant.

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##### Home Assignment 2 / Re: Home Assignment 2 Problem 2(a)
« Last post by Zhiman Tang on January 18, 2019, 09:17:57 PM »
I think you need to find the value of u(0,0), below is my answer, but idk whether it is correct or not.
consider u(x,y) = f(x/y)
in this case, x=0 is not a problem, but y=0 is a problem
we can write u(0,y) = f(0),
to make u a contiunuos solution, we need u(0,0)=f(0) since the value of u can't jump at y=0
so u(0,0)=f(0) is the condition to make it continuos.
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##### Home Assignment 2 / Re: Secondary Textbook Chp1.2 Exercise 10
« Last post by Victor Ivrii on January 18, 2019, 09:05:21 PM »
Read Section 2.1, Subsection  "semilinear equations"
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##### Home Assignment 1 / Re: problem 4 (25)
« Last post by Wanying Zhang on January 18, 2019, 08:04:40 PM »
I would start from the beginning.
$$\frac{u_{xy}}{u_x} = \frac{u_x u_y}{u_x} \Rightarrow \frac{u_{xy}}{u_x} = u_y$$
Integrate both sides,
$$\ln{u_x} = u + f(x)$$
$$u_x = e^{u+f(x)} = e^u \cdot g(x)$$
$$\frac{\partial u}{\partial x} = g(x)e^u$$
$$\frac{\partial u}{e^u} = g(x)\partial x$$
Integrate both sides,
$$-e^{-u} = G(x) + h(y)$$
$$u(x,y) = -\ln (-G(x) - h(y))$$
OK. V.I.
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##### Home Assignment 2 / problem1 a (5)
« Last post by Zhiman Tang on January 18, 2019, 07:26:28 PM »
the integral curve is： dt = dx/x
if I integrate both sides, I get t - lnx = c, so the general solution is u = f(t-lnx)
however, it I move x to the left side: xdt = dx, and integrate both sides, I get xt - x = c. You cannot do this, since in ODE , describing integral curves, $x$ and $t$  are not independent. V.I.

the general solution becomes u = f(xt-x)
I feel the second approach is wrong, but I cannot tell where did I do wrong.

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##### Home Assignment 1 / problem 4 (25)
« Last post by Zhiman Tang on January 18, 2019, 07:01:19 PM »
uxy = uxuy
I use the hint and divide both sides by ux. I get ux = eu * f(x). I am stuck there. Could anybody help me out?
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##### Home Assignment 2 / Secondary Textbook Chp1.2 Exercise 10
« Last post by Wanying Zhang on January 18, 2019, 06:16:16 PM »
I have difficulties solving this problem: $u_x + u_y + u = e^{x+2y}$. I get $\frac{dy}{dx} = 1$ from the characteristic equation and have no ideas what should be followed. Any reply would be appreciated.
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