How Are the Roots of t^2-t-2 Related to the Solutions of y''-y'-2y=0?

[gigi9]

Someone please help me to do this problem below. Thanks
1)a) Verify that y= e^(-x) and y= e^(2x)are both solutions of the differential equation y"-y'-2y=0
b) Factor the polynomial t^2-t-2. What connection do you see between the roots of this polynomial and the solutions of y"-y'-2y=0 ?---What does "roots" means in this case??..I don't understand.
*** t^2-t-2=(t-2)(t+1)

 

[HallsofIvy]

Once again, this really should be in homework (its a territorial matter!) and you really should show what you have tried.

However I will give you a hint: To show that y= e^-x and y= e^2x are solutions to the differential equation y"-y'-2y=0 means to find y' and y" for each function, put them into the given equation and see what happens!

") Factor the polynomial t^2-t-2. What connection do you see between the roots of this polynomial and the solutions of y"-y'-2y=0 ?---What does "roots" means in this case??..I don't understand."

I started to write "exactly what it means in algebra" but you do have a point (perhaps inadverdant): Strictly speaking to talk about "roots" you must have equation, not just a polynomial. This problem should have said either "the roots of the equation t²- t- 2= 0" or "the zeroes of the polynomial t²-t-2".

Okay, you have factored t²- t- 2= (t+ 1)(t- 2). Now WHAT are the roots of t²- t- 2= (t+ 1)(t- 2)= 0 and what possible connection could that have with e^-x and e^2x?

 

[M98Ranger]

To factor the polynomial t^2-t-2, we can use the standard method of grouping or the quadratic formula. Let's use the grouping method:

Step 1: Write the polynomial in the form t^2+at+b. In this case, a=-1 and b=-2, so we have t^2-t-2.

Step 2: Find two numbers that when multiplied give you the constant term (b) and when added give you the coefficient of the middle term (a). In this case, the numbers are -1 and 2, since (-1)(2)=-2 and -1+2=1.

Step 3: Rewrite the polynomial using these two numbers as coefficients for the middle term. t^2-t-2 can be rewritten as t^2+2t-t-2.

Step 4: Group the first two terms and the last two terms together. We get t(t+2)-1(t+2).

Step 5: Factor out the common term (t+2). We get (t+2)(t-1).

Therefore, the polynomial t^2-t-2 can be factored as (t+2)(t-1).

Now, let's look at the connection between the roots of this polynomial and the solutions of y"-y'-2y=0.

In the polynomial t^2-t-2, the roots are the values of t that make the polynomial equal to zero. In this case, the roots are t=-2 and t=1.

In the differential equation y"-y'-2y=0, the solutions are the values of y that make the equation true. In this case, the solutions are y=e^(-x) and y=e^(2x).

Notice that when we substitute t=-2 in the polynomial, we get (t+2)(t-1)=(-2+2)(-2-1)=0, which means that t=-2 is a root of the polynomial. Similarly, when we substitute t=1 in the polynomial, we get (t+2)(t-1)=(1+2)(1-1)=0, which means that t=1 is also a root of the polynomial.

This shows that the roots of the polynomial t^2-t-2 are directly related to the solutions of the differential equation y"-y'-2y=0. This is because both the polynomial and the differential equation have the