Xt xo ut bt letting t 0 and using the fact that xt xo x when...

Xt = xo + ut + Bt Letting t → 0 and using the fact that Xt ~ Xo = x when t is small, we arrive at x + Gx = 0. (0.1) This is the ODE that v should satisfy. iv. What you need to do: Solve the ODE (0.1). To solve an ordinary differential equation like this, we need to find a particular solution and the general solution. Try a particular solution of the form fp(x) = Ax + B where A and B are some constants that you will need to determine. The general solution should be of form fg(x) = C1e^(B1x) + C2e^(B2x) The coefficients C1 and C2 are free constants that we will determine in the next part, but you should be able to determine B1 and B2 by solving the associated characteristic equation (a quadratic equation). Therefore, your final answer for this part should be of form v(x) = fp(x) + fg(x) = Ax + B + C1e^(B1x) + C2e^(B2x) where A, B, 1, 2 are constants that you can determine. V. What you need to do: Use the appropriate growth condition to argue that C1 = C2 = 0. Suppose it is given that v(x) grows at most linearly when x as x → 0 (see Remark 0.1). Use this to argue C1 = C2 = 0. You are not asked to write a 1-million-page-proof, just one or two sentences to explain your line of thoughts are good enough vi. The solution v(x) = Ax + B you get here should be the same as the solution you get from Part (a). v(xo) X0 a^2

Transcript

00:01 In this problem we have given differential equation y double dash plus y -dase minus xy equal 0 so in the first part it is given y equal t to power rt it is given that the equation has the solution of the form y is equal to power rt so we have to find the all values of r which provides the solution to this differential equation first.

00:36 Now further we'll solve differentiate it and y double dash is r square to power r t substituting these values in this differential equation we will get y double dash plus y -dash minus 6 0 so this will give r square 8 to power r t plus r to power r t minus 6 to power r t equal 0 taking it for r t outside r square plus r minus six is zero since it for r t is non -zero hence r square plus r minus six equal zero so this is called characteristic equation characteristic equation and the solution of this characteristic equation give the values of r that will form the solution of the given differential equation so to solve this equation here factor is 3r minus 2 r taking r outside so it will give two factor of this quadratic equation so taking r plus 3 equal 0 we will get r equal minus 3 and r minus 2 and r minus 2 we will get r equal to so here let's take first value r1 equal minus 3 and this is r2 2 we have two values of r which give the solution of this ordinary differential equation since we have assumed that y equal e to power r t solution so far r1 y1 equal e to power r1 t solution that is y1 equal e to power minus 3 t solution and similarly y2 equal e to power r2 is a solution that is y2 equal 8 to power 2 t here in the second part it is given that if for a differential equation if y1 is a solution and y2 is a solution then y1 plus y y2 is also a solution and c y t is also a solution so it is given that for a differential equation the general solution can be written as y t y g t equals c1 e to r1 t plus c2 to power r2t so plug here these values the general solution is c1 and this is y1 it for r1 t here we have taken 8 to minus 3 t and similarly plus c2 8 to be it for 2 t so this is the general solution of the given differential equation y double dash plus y dash minus 6y equal 0 now we have here initial conditions y 0 equal y0 and y dash 0 equal v0 now we have to using this initial condition we have to solve and finding the values of c1 and c2 in the terms of y and v0 so putting here using y0 equal y0 in this solution so we will get here y0 equal c1 8 to power 0 plus c2 8 to power 0 so it will give y0 equal c1 plus c2 that is equation 1 let's now so y -d as t differentiating this general equation we will get here minus 3 c1 e to power minus 3 t plus 2 c2 e to power 2 t and substituting the initial condition y -d s 0 equal v0 this one we will get here y -d s 0 equal minus 3 c1 e to power 0 plus 2 c2 e to power 0 so this will give us v0 equal minus 3 c1 plus 2 c2 so this is equation 2 now we will solve equation 1 and 2 for the value of 7 and c2 so let's multiply 3 in this equation and add in this equation second so 3 y0 equal 3 c1 plus 3 c2 so adding both we'll get here v0 plus 3 y0 equal this will cancel out and here 5c2.

06:28 So we will have here 3y0 plus v0 equal 5c2 that is c2 equal 3y0 plus v0 by 5...

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