Thread: 4y''-4y'+4y=0

Find the r value (which is a constant)

y = e^rx

solution of the following equation
y''-4y'+4y=0

I got

y'' = e^rx + (r^2)(e^rx)

I plugged in y' and y'' and y to the original equation and cannot get an answer out of it after trying to a very long time!

I would first divide through by 4 just to simplify the differential equation, then compute the first and second derivatives and plug them into the equation.

What do you find?

OOPS!~ I wrote the problem down wrong.

Its y''-4y'+4y=0

Sorry!

Okay no worries, just forget about dividing through by 4, but the rest still applies. What do you get when you find the first and second derivatives of the given solution $\displaystyle y=e^{rx}$, and plug them into the equation?

You have y = e^(rx) Thus we get y'=re^(rx) and y"= r^2 e^(rx)
On substituting the values in given equation we get
r^2 e^(rx) -4re^(rx) + 4re^(rx) = 0
That leads to
e^(rx) [ r^2- 4r + 4 ] = 0 OR e^(rx) ( r-2)^2 = 0
now you can take it on from here

Originally Posted by ibdutt

You have y = e^(rx) Thus we get y'=re^(rx) and y"= r^2 e^(rx)
On substituting the values in given equation we get
r^2 e^(rx) -4re^(rx) + 4re^(rx) = 0
That leads to
e^(rx) [ r^2- 4r + 4 ] = 0 OR e^(rx) ( r-2)^2 = 0
now you can take it on from here

[rant]
What's left for the OP to do really?

Did you not notice I was trying to get the OP to do the problem on their own with minimal assistance? Then you come along and work 95% of the problem and leave the OP to simply make the conclusion. In trying to get the OP to learn by doing, we at MHF have now failed, and my time was simply wasted.

Please do not "snipe" me or others like this in the future...it is bad forum etiquette. If you are the first to help, then if you want to be lazy and simply work the problem for them, that is your call, albeit a bad one. But to come along after others are actually making the effort to teach rather than simply demonstrate an ability to work the problem is just in poor taste, and I know from experience that I am not the only one that does not appreciate this at all.

I used to naively post full solutions until I realized how lazy and ineffective this actually is, but I never did this after others had already posted help. It just seems to me to be common sense that this is inconsiderate.

Also, if you are going to help people here, it would be good for you to make the effort to learn to use $\displaystyle \LaTeX$ so that the help you provide is easy on the eyes.

[/rant]

It's okay - maybe he did not realize that that was proper etiquette. We only have a short time here on earth lets all be peaceful =)

Dear MarkFl, Please learn to be polite. Each teacher has own way to communicate. It is not face to face communication. One cannot keep passing on point by point that is what I think. Anyhow thanks for the advice.