Question
Problem 3 points) For u(x;t) defined on the domain of 0 <x < 21 and 20, solve the PDE_du d2u du (1 +t) 1 41 dt dx2 dxwith periodic boundary conditions in the x-direction; and the boundary condition in the t-direction given asu(x; 0) = 1 sin(2x) cos(2x)We expect closed-form solution which consists of only finite number of terms and without any unevaluated integrals. The solution should be expressed in real functions
Problem 3 points) For u(x;t) defined on the domain of 0 <x < 21 and 20, solve the PDE_ du d2u du (1 +t) 1 41 dt dx2 dx with periodic boundary conditions in the x-direction; and the boundary condition in the t-direction given as u(x; 0) = 1 sin(2x) cos(2x) We expect closed-form solution which consists of only finite number of terms and without any unevaluated integrals. The solution should be expressed in real functions
Answers
Find the particular solution of the differential equation that satisfies the boundary condition.
Differential Equation $\qquad$ Boundary Condition
$y^{\prime}+y \tan x=\sec x+\cos x \quad y(0)=1$
Okay, So for party, we have that. Why of zero equals two implies that C one is equal to And why have pie over to equal zero implies that C two is equal to zero. So then we can write why of X is equal to co sign of X And this is our unique solution our unique solution. So this is part of a no part B. When we use the same kind of approach, we will get that Why was your equals? To imply I see one equals two again. But why have pi equals zero implies that C two is equal to Well, nothing because actually, the equation will vanish. So we'll get zero equals too coarse and pie plus c two sign pi So see to sign pi is C two times zero which is zero to see too vanishes and you got zero equals negative too, which is, um not true. So we have no solution for this part. And for part c, we have C one equals two and we will obtain see two equals two is well, actually sorry, I take that back. We don't get seat equals two. We get negative two equals negative too. And this is coming from the initial condition. Why have pie is equal to negative too? But this is true no matter what. So there are infinitely many solutions for part C infinitely many solutions.
Mhm. Right. So in this problem were given the differential equation to find as x squared -2 sign effects. So the first thing I would do is just think about what is the anti directive. So you have to add one to your exponent, multiply by the reciprocal. And here's the thing that a lot of students get confused on that. The anti drift of negative sign would actually be positive co sign. And the reason why I know that off the top of my head is that the drift of of co sign is negative sign. And that too just goes along for the ride. Um So this is the correct anti derivative. And now what we need to establish is that when X. Is pie, The why value is zero. So what we have to do is is plug in zero for why plug and pie for your exes and figure out what you're see. Value would be Now just a reminder if you look at your unit circle, co sign of pie pies over here and co sign is negative one. So the C. Value when I go to solve this um uh This would be negative too because two times negative on his name too. So I would add two to the left side and then subtract that pi cubed over three. Um And what I need to do is go back to my value for C And rewrite that equation y equals 1 3rd X cubed. Was it plus to co sign of X. Yeah. Uh And then the C value replace it with that two minus pi cubed over three. Um And this is your correct answer because the derivative of this is the differential equation appear and it satisfies the initial condition that one excess pie, The Y Value is zero.
Today we're going to solve from the number 17 here The given equation is valued eyes In due course, the square it's plus y minus one equals zero. So why of zero? It was fighting. So why you guys plus one by caused square X in the white equals one by cause the square x here p off Mexico's one because square it's cure fixes equal to one day called square X So integrating factor recalls he integral six square x the X which is equal to us to Panix Why in do it is too planets it was integral into the power Panix in Duke six correx the X right into into the power. Then it's, you know, equals into the power on its plus c. So eat the power Panix into white in tow. The par minus 10 x This one lets see u to the power minus panics. All right, so boundary condition is gonna like y of zero equals five so we can find constancy value for file because one plus see it, the bird minus 10 0 so five equals one plus e in the one see because five minus one, which is equal to four. So Y equals one plus four. Developed by into the power dynamics, which is the particular solution. Thank you.
Hello, everyone. Today we're going to solve problem number 20 here The U. N first order differential location is wide eyes plus two X minus one in the white equals zero. This is P O effects and this is que all fits so integrating factor equals e in the girl two X minus one The it's it is going to eat the products in the X minus one. So why India Integrating factor equals integral cure Fix Indu integrating factor that ISS by indu integrating factor It was individual cure fits into the X So what you'll be getting is like Why in tow bar X into X minus one equals C Well, I equals c different. But you know the products in the X minus one border conditions are like way off one equals toe two equals C diverted by you to the ball one Indo zero So Sequels what he calls to divided by you Do the projects in tow X minus one. Thank you