You are given general sol'n - find differential equation

[accountkiller]

Homework Statement

You are given a solution - find the homogeneous, 2nd order differential equation ay'' + by' + cy = 0

Homework Equations

y(x) = c₁ + c₂e^-10x

The Attempt at a Solution

I found the derivatives first.
y'(x) = c₁ - 10c₂e^-10x
y''(x) = c₁ + 100c₂e^-10x

But I have no idea where to go from there. The answer is y'' + 10y' = 0. Looking at that answer, yes I see why that is the answer, but what work did I have to show to get to that answer?

I'd appreciate any guidance, thanks!

 

[Mark44]

Homework Statement

You are given a solution - find the homogeneous, 2nd order differential equation ay'' + by' + cy = 0

Homework Equations

y(x) = c₁ + c₂e^-10x

The Attempt at a Solution

I found the derivatives first.
y'(x) = c₁ - 10c₂e^-10x
y''(x) = c₁ + 100c₂e^-10x

For starters, the derivative of any constant (e.g., c₁) is 0.

But I have no idea where to go from there. The answer is y'' + 10y' = 0. Looking at that answer, yes I see why that is the answer, but what work did I have to show to get to that answer?

I'd appreciate any guidance, thanks!

 

[Mark44]

From the solution you are given, e^0x (= 1) and e^-10x are basic solutions. These have something to do with the characteristic equation.

 

[accountkiller]

Ah, duh! Right, constants ' = 0, so y'(x) = -10c₂e^-10x and y''(x) = 100c₂e^-10x.

So I understand that e^0x (=1) and e^-10x are solutions. But how do I involve the characteristic equation?
ar² + br + c = 0

How do I find out the a, b, c? Are they those particular solutions? So a = e^0x (=1) and b = e^-10x and c = ... ? 0?

 

[accountkiller]

Okay, about the characteristic equation..
Since the book says we are supposed to guess y₁ = e^r₁x and y₂ = e^r₂x, that means my r₁ = 0 and my r₂ = -10.

Now.. where to go from there?

 

[Mark44]

OK, good. These are the roots of the characteristic equation ar² + br + c = 0. And a, b, and c are the coefficients of the diff. equation ay'' + by' + cy = 0. Can you continue from here?

 

[HallsofIvy]

If r_1 and r_2 are roots of a quadratic equation, then the equation is of the form a(x- r_1)(x- r_2)= 0 for some number a.

 

[accountkiller]

Ah, so if r₁ = and r₂ = -1 are the roots of ar² + br + c = 0, then I have (r + 0)(r + 10) = 0 so my characteristic equation is r² + 10r = 0, so (1) and (10) are my a and b constants, so putting it in ay'' + by' + cy = 0 gives me the differential equation (1)y'' + 10(y') = 0.

I'm usually good at math but this chapter has, for some reason, been just difficult for me to 'click' with whatever I'm supposed to be understanding. Thanks for all of your help Mark, I really appreciate it!