Consider the differential equation y” + 2y' + y = 0. (a)...

Consider the differential equation y'' + 2y' + y = 0. (a) Find r1, r2, roots of the characteristic polynomial of the equation above. r1, r2 = -1, -1 (b) Find a set of real-valued fundamental solutions to the differential equation above. y1(t) = e^-t y2(t) = e^-t (c) Find the solution y of the the differential equation above that satisfies the initial conditions y(0) = 1, y'(0) = 1. y(t) =

Transcript

00:01 Okay here we have a given differential equation and we want to find the general solution right so first of all we need to write the auxiliary equation right so here auxiliary equation is a r square plus 2 r plus 1 is equal to 0 so here we can write r plus 1 rest 2 is equal to 0 which means r plus 1 is equal to 0 therefore r is equal to minus 1 with multiplicity of 2 right so our root r1 and r2 that is equal to minus 1 and minus 1 right so your first answer is absolutely correct now for the one real root the general solution is given by y is equal to c1 multiplied by e -d s 2 r plus c2 multiply by t d 2 er s 2 r and here r that is equal to minus 1 sorry here we need to multiply by t we we forgot that so here we can write a c1 multiply by e -d -2 minus t plus c2 multiply by t multiply by e -d minus t right so our first root is y1 is equal to e -d2 minus t and the second route is t multiplied by e -d to minus t right so this is your answer number b right okay now the initial condition is y of 0 is equal to 1 right which means we need to substitute t is equal to 0 and y is equal to 1 in this equation right so here er s to 0 that is equal to 1 so 1 multiply by c1 that is equal to c1 and here 0 multiply by any value that will be 0 therefore for c1 is equal to zero sorry c1 is equal to 1 not 0 right because you're on the left side we have a 1 right so here c1 is equal to 1 now substitute the value of c1 in this equation now we need to take a derivative now derivative of e -d s 2 minus t is a minus e -d s 2 minus t and for the second solution we need to use the product rule for the derivative right so here first function will be as it is multiply by derivative of the second function plus second function will be as it is multiply by derivative of the first function right now simplify this okay now the second initial condition is uh y -dhs of zero is equal to one so here we need to substitute t is equal to zero and y -dase is equal to one e -dase to zero...

Question

Please give Ace some feedback