Question

if A and B are constants , then what will be the solution space of the D.E y"-8y'+16y=0 ?

Answer 1

Have you covered finding solutions of general equations using the characteristic equation?

In this case, you could substitute r for every derivative of y and set to 0, which would give you:

\[r^2 - 8r + 16 = 0\]

If you solve this equation, you will find that
\[(r-4)^2=0\]

This means you have the 'repeated solutions' of r = 4 and r = 4.

Normally, the solution would consist of
\[y=c_{1}e^{rx} + c_{2}e^{r} + ... + c_{n}e^{rx}\]

with a separate term for each value of r.

However, since you have a repeated root of r = 4 (that is, the two solutions for r that are the same), you must insert an extra 'x' in one of the terms. (Because you cannot have two terms that are exactly the same; either the exponent must be different, or you need a different leading power of x.) So instead of
\[y=c_{1}e^{4x} +c_{2}e^{4x}\]
You need
\[y=c_{1}e^{4x} +c_{2}xe^{4x}\]

Answer 2

why did you insert an x into the second term.?

Answer 3

Hi there,
If I had not inserted an X into the second term, the two terms would be equivalent. (Since c1 and c2 are just some constant to solve for, they would combine into a single 'c')

I don't have the proof with me, but it is possible to prove that when you have repeated roots, where lambda is the same (any time you have (r-b) to a power greater than 1), you must insert a power of X to make the terms each unique.

Another example: say I had (r-5)^3:
Then I would have a repeated root of r=5.

So my terms would be c1*e^(5x)+ c2*(x)*e^(5x)+c3*(x^2)*e^(5x)

I like to just think of each power of X as necessary to make each term unique. If you want a proof for this rule, try googling 'repeated roots proof'.

Sorry my equation formatting is a little strange; I am answering from my mobile.