Need help with Differential equation

  • Thread starter James889
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  • #1
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Hi, i would like some help with his one

I have
[tex]\left\{\begin{aligned}y^{\prime\prime}+10y^{\prime}+25y = 0 \\
y(1) = 0 \\
y^{\prime}(1) = 2\end{aligned} [/tex]

So the first part is simple

The general solution is on the form
[tex]-5\pm\frac{\sqrt{10^2-4\cdot25}}{2}[/tex]

it's a double root
So we have
[tex] y = Ae^{-5t}+Bte^{-5t}[/tex]

And
[tex]y^{\prime} = -5Ae^{-5t} -5Bte^{-5t}[/tex]

[tex]-5e^{-5t}(A+Bt)[/tex]
And since t = 1

[tex]A=e^5~~ B=-e^5[/tex] works for the first condition but not for the second.

Hm, how do you do this?
 
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Answers and Replies

  • #2
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you need to differentiate the GS and plug y=1 into it.
y'=B.exp(-5t) - 5(A+Bt).exp(-5t)
Since A+B=0,y'(1)=B.exp(-5)=2
so B=2.exp(5) and A=-2.exp(5)
 
Last edited:
  • #3
192
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you need to differentiate the GS and plug y=1 into it.
y'=B.exp(-5t) - 5(A+Bt).exp(-5t)
Since A+B=0,y'(1)=B.exp(-5)=2
so B=2.exp(5) and A=-2.exp(5)

Huh?

Are you sure?

Isn't the derivative of [tex]y = Ae^{-5t}+Bte^{-5t} \rightarrow y^{\prime} = -5Ae^{-5t}-5Bte^{-5t} ? [/tex]
 
  • #4
vela
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Two words: product rule.

Check your expression for [tex]y'[/tex].
 
  • #5
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Huh?

Are you sure?

Pretty sure....

Isn't the derivative of [tex]y = Ae^{-5t}+Bte^{-5t} \rightarrow y^{\prime} = -5Ae^{-5t}-5Bte^{-5t} ? [/tex]

No, the derivative of A.exp(-5t) is -5A.exp(-5t) and the derivative of Bt.exp(-5t) is -5Bt.exp(-5t) + B.exp(-5t)
 
  • #6
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ok, thanks.
Sigh... i must be the worst math student ever...
 
  • #7
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Cheer up :) It's an easy mistake to make. The main point is that you got the main method right. If I had a pound for the number of times I've made mistakes like that I'd be very rich ! See my thread on implicit differentiation if you want a laugh !
 
  • #8
HallsofIvy
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Your problem was that you didn't use the product rule on [itex]te^{-5t}[/itex].
 
  • #9
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Hey,

I'm going to use this thread, because I was checking it out (studying differential equations too) and I didn't quite understand what he did in the beginning, how James got to those 2 equations about y and y'.

The way I would have done this is set p = y', therefore p' = dp/dx = p·dp/dy and then it becomes dp/dy + 10 = -25·y·p-1 which is a Bernouilli equation.

Thanks for your help too!
 
  • #10
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Hey,

I'm going to use this thread, because I was checking it out (studying differential equations too) and I didn't quite understand what he did in the beginning, how James got to those 2 equations about y and y'.

James' equation for y is just the (explicit) general solution which follows from the solution of the characteristic equation (repeated real root). His equation for y' should have just been the derivative but he forgot to use the product rule for the term involving the constant B.

The way I would have done this is set p = y', therefore p' = dp/dx = p·dp/dy and then it becomes dp/dy + 10 = -25·y·p-1 which is a Bernouilli equation.

Thanks for your help too!

Sorry, but I don't follow......
 
Last edited:
  • #11
vela
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When you have a linear differential equation with constant coefficients, you can take advantage of the fact that the exponential function is the eigenfunction of the derivative operator, D:

[tex]D e^{rx} = r e^{kx}[/tex]

You assume a solution of the form [tex]y=e^{rx}[/tex], substitute it into the differential equation, and solve for [tex]r[/tex]. In this particular case, you'd get

[tex](r^2+10r+25)e^{rx}=0[/tex].

Since the exponential is never zero, the polynomial factor in front must be zero. If you get two distinct roots, [tex]r_1[/tex] and [tex]r_2[/tex], you have your two linearly independent solutions, [tex]y_1 = e^{r_1 x}[/tex] and [tex]y_2 = e^{r_2 x}[/tex], for this second-order differential equation. In this case, however, there was a repeated root. To get the second independent solution, you have to multiply by [tex]t[/tex], so [tex]y_2 = te^{rx}[/tex], and the general solution would be [tex]y=A e^{rx} + B te^{rx}[/tex].
 

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