Question

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

Answer 1

@pooja195

Answer 2

@ganeshie8

Answer 3

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

Answer 4

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

Answer 5

Now you are able to finish it.

Answer 6

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?

Answer 7

Oops, x, not z

Answer 8

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

Answer 9

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.

Answer 10

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

Answer 11

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

Answer 12

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).

Answer 13

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.

Answer 14

values of x*

Answer 15

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

Answer 16

Yes, you are right.

Answer 17

0 and 4?

Answer 18

Yeah. Spot on!

Answer 19

Wow, that was easier than I expected

Answer 20

Thanks Chill!

Answer 21

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.

Answer 22

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

Answer 23

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

Answer 24

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

Answer 25

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

Answer 26

Sure. Ask away.

Answer 27

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

Answer 28

Show me what you have done until now.

Answer 29

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

Answer 30

Here's what I did

Answer 31

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

Answer 32

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

Answer 33

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

Answer 34

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

Answer 35

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

Answer 36

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

Answer 37

But yes, you did it right.

Answer 38

Huh... Okay, well thanks again man!