Find the two values of k for which y(x) = e^(kx) is a solution of the differential equation y" - 4y' + 0y = 0

- anonymous

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- anonymous

@pooja195

- anonymous

@ganeshie8

- ChillOut

All right, first we should have \(k²*e^{kx}-4ke^{kx}=0\)

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- ChillOut

which leads to \((k-4)*k*e^{kx}=0\)

- ChillOut

Now you are able to finish it.

- anonymous

Woah woah woah, diff eq's aren't my strong suit, I'm just getting started. could you walk me into getting the k^2 * e^(kz)-4ke^(kx) = 0?

- anonymous

Oops, x, not z

- anonymous

Is it just doing second derivative of the function, and the first derivative, then subbing into the form given?

- ChillOut

All right. For differential equations you must know derivatives... The differential equation in question \((y''(x)-4y'(x)+0y=0\)) is a second order linear ordinary differential equation,
y''(x) is the second derivative of y(x) and y'(x) is the first derivative.

- ChillOut

All I did was the differentiation of the given solution. Afterwards I plugged the solution in the equation.

- anonymous

So with this all plugged in now, I want to solve for x = 0? Or find k in terms of x?

- ChillOut

No, you don't want to solve in terms of x. You want find a value for k in which the left-hand side is equal to the right-hand side. (This is stated in the beggining of the question).

- ChillOut

The question doesn't asks for values x, and most likely will never ask, unless it is a boundary value problem or initial value problem.

- ChillOut

values of x*

- anonymous

So I just want values for k that will make \[ke ^{kx}(k-4)\] equal to zero? I'm sorry if I'm still a little confused

- ChillOut

Yes, you are right.

- anonymous

0 and 4?

- ChillOut

Yeah. Spot on!

- anonymous

Wow, that was easier than I expected

- anonymous

Thanks Chill!

- ChillOut

When it gives you a solution, you will want to plug that solution and its derivatives in the ODE. That's how this kind of question is done.

- anonymous

ODE being Order of Diff Eq's? Will the form given in this problem always be the same?

- anonymous

Meaning y" - 4y' + 0y = 0

- ChillOut

Nope, Ordinary Differential Equation. We have the PDEs too, but that's later for you :P

- anonymous

Can I ask one more question on this topic? I've done out the work but can't see where Differential Equations fit in

- ChillOut

Sure. Ask away.

- anonymous

The solution of a certain differential equation is of the form \[y(t)=ae ^{2t} + be ^{3t}\] where a and b are constants. The solution has initial conditions y(0) = 4 and y'(0) = 3. Find the solution bu using the initial conditions to get linear equations for a and b

- ChillOut

Show me what you have done until now.

- anonymous

I did it out, and I'm pretty sure that a = -1 and b = 5. I thought I was supposed to thenn just substitute those numbers in for a and b, and call it a day, but my homework program says that's wrong

- anonymous

Here's what I did

- anonymous

f(0)=4, so a+b should =4, and we can get a = 4-b

- anonymous

Then I found the derivative, y'(t) = 2ae^(2t)+3be^(3t)

- anonymous

Subbed in 0 and set it equal to 3, where I found that 3 = 8-b, so b = 5

- anonymous

Then I went back and subbed the new b into a = 4-b to get a = -1

- anonymous

Submitted as an answer -1e^(2t)+5e^(3t) but webwork said it was wrong

- ChillOut

Well, I dunno how they accept the answers, so in this one I can't help you :(

- ChillOut

But yes, you did it right.

- anonymous

Huh... Okay, well thanks again man!

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