An initial-value problem is given, involving a second order homogeneous linear d
ID: 2942472 • Letter: A
Question
An initial-value problem is given, involving a second order homogeneous linear differential equation, with a pair of functions y1(x) and y2(x). Verify that these two functions are particular solutions to the given differential equation, and then find a linear combination of these solutions that satisfies the given initial value problem.1) I.V. Problem: y''-4y'+4y=0 with y(0)=1 and y'(0)=6
Functions: y1(x)=e^2x and y2(x)=xe^2x
[If it helps, the answer in the back of the book is y(x)=e^2x+4xe^2x]
Explanation / Answer
I'm going to talk through most of this so please bear with me: to determine if a function is a solution to the DE, find the necessary derivatives of the said functions. In this case, y1(x)=e^(2x) y1'(x)=2e^(2x) y1''(x)=4e^(2x) substitute into the DE: 4e^(2x)-4(2e^(2x))+4(e^(2x))= 0. since it satisfies the DE, it is a solution. It is important to notice here that if you added an arbitrary constant to y1, so that y1(x)=ce^(2x), when you take the derivatives, and plug them in, the solution will still hold. This is what we refer to as a family of solutions. So do as above to confirm that y2(x) is also a solution (the derivatives are a little uglier though) We also have a really nice theorem (probly in your book somewhere if you wish to confirm) that says "we can add two solutions of a DE and get a new, bigger one!" Since we have established that the two solutions are still solutions if we attach a constant, adding them together yields: y1(x)+y2(x)=(c1)e^(2x)+(c2)xe^(2x) = Y //the bigger solution(confirm if you wish) Now for the initial value stuff: Y(0)=1 gives (c1)e^(2(0))+(c2(0)e^(2(0))=c1=1. and Y'(0)=6 gives: 2e^(2(0))+(c2)e^(2(0))+2(0)e^(2(0))=6 which implies that c2=4. So our particular solution is Y(x)=e^(2x) + 4xe^(2x).