Question

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I don't know do you find the characteristic form?

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For this theres a typo it should be a+d <0 is stable and a+d >0 is unstable

How do you start with these proofs? I know that if the eigenvalues are both negative then its stable and if there both positive its unstable. And if one is negative and the other is positive then its unstable saddle. How do you put that in proof form?

Answer 1

Since you are given that y1 and y2 are solutions to the equation ay''+by'+cy=0, you know that:\[ay_1''+by_1'+cy_1=0\]and\[ay_2''+by_2'+cy_2=0\]Does this make sense?

Answer 2

yes

Answer 3

You need to show that if those two statements hold, then:\[a(c_1y_2+c_2y_2)''+b(c_1y_2+c_2y_2)'+c(c_1y_2+c_2y_2)=0\]You can do this by expanding the left side out, and using the first two statements.

Answer 4

I don't get what you did there

Answer 5

To show that a function is a solution to the equation ay''+by'+cy=0, you need to brute force plug it into the equation and show you get zero.

The question is saying, if y1 is a solution, and y2 is a solution, show that c1y1+c2y2 is also a solution. You now have a new function c1y1+c2y2, and you need to plug it into the differential equation to show it comes out to 0.

Answer 6

You start by taking the first and second derivative of c1y1+c2y2:\[(c_1y_1+c_2y_2)'' = c_1y_1''+c_2y_2''\]Similarly for the first derivative. Bah i noticed a little typo, it should be c1y1+c2y2 in that equation.

Answer 7

Yeah I was going to say

Answer 8

Basically you get this:\[a(c_1y_1+c_2y_2)''+b(c_1y_1+c_2y_2)'+c(c_1y_1+c_2y_2)=\]\[a(c_1y_1''+c_2y_2'')+b(c_1y_1'+c_2y_2')+c(c_1y_1'+c_2y_2')\]Next distribute the coefficients out, rearrange the terms, and it should be clear what to do after that.

Answer 9

is the c term suppose do be primed?

Answer 10

ack its not >.< wow i need to proof read my posts >.< sry about that.

Answer 11

So ac1y1'' + bc1y1' + cc1y1 = 0
ac2y2'' + bc2y2' + cc2y2 = 0
so for the first one the solution is c1y1 and the second one is c2y2, but the solution should be c1y1 + c2y2?

Answer 12

Right. We see that c1y1+c2y2 is a solutions since:\[a(c_1y_1′′+c_2y_2′′)+b(c_1y_1′+c_2y_2′)+c(c_1y_1+c_2y_2)\]\[=ac_1y_1′′+ac_2y_2′′+bc_1y_1′+bc_2y_2′+cc_1y_1+cc_2y_2\]Regrouping like you have in your post yields:\[(ac_1y_1''+bc_1y_1'+cc_1y_1)+(ac_2y_2''+bc_2y_2'+cc_2y_2)\]\[c_1(ay_1''+by_1'+cy_1)+c_2(ay_2''+by_2'+cy_2)\]Since we were given that both y1 and y2 were solutions, we know the terms inside the parenthesis evaluate to 0, so we obtain:\[c_1(ay_1''+by_1'+cy_1)+c_2(ay_2''+by_2'+cy_2)=c_1\cdot 0+c_2\cdot 0 = 0.\]That shows c1y1+c2y2 is also a solution.

Answer 13

Ok that makes sense thanks