How to find verify solution to differential equation

  • Thread starter digipony
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  • #1
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Homework Statement



See attached photo

Homework Equations





The Attempt at a Solution


I don't even know where to start on this problem. Usually, if I am given a possible solution to a DE, say y(t)=cos(t) as a solution to y''-y'+y=sin(t), I find whether it was a solution or not by manipulating it and then plugging it into the second equation that it may be a solution to and then seeing whether the resulting statement is true. For the ex. problem I give in this section, I would do this:
y'= -sint
y''= -cost
then sub them in: -cost +sint +cost = sint
and by simplifying get: sint = sint
so for this case, it y=cost is a solution.
But with the problem that is in the attached photo I have no idea what to do.

Does anyone have an guidance as to how I can solve the attached problem?
Thanks :)
 

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Answers and Replies

  • #2
SammyS
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Homework Helper
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Homework Statement



See attached photo

Homework Equations



The Attempt at a Solution


I don't even know where to start on this problem. Usually, if I am given a possible solution to a DE, say y(t)=cos(t) as a solution to y''-y'+y=sin(t), I find whether it was a solution or not by manipulating it and then plugging it into the second equation that it may be a solution to and then seeing whether the resulting statement is true. For the ex. problem I give in this section, I would do this:
y'= -sint
y''= -cost
then sub them in: -cost +sint +cost = sint
and by simplifying get: sint = sint
so for this case, it y=cost is a solution.
But with the problem that is in the attached photo I have no idea what to do.

Does anyone have an guidance as to how I can solve the attached problem?
Thanks :)
Here's the "photo":

attachment.php?attachmentid=50458&d=1346730413.jpg


Well, find [itex]\displaystyle \frac{\partial}{\partial x} f(x-ct)[/itex] and then the partial with respect to x of that.

Do the same with partials with respect to t.

Then plug all that in.
 
  • #3
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I don't have much experience with partial derivatives, am I on the right track here: the first partial derivative with respect to x, [tex]\displaystyle\frac{\partial}{\partial x} f(x-ct)[/tex] be: 1, therefore the second would be 0?
 
  • #4
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Sorry for the formatting on that, I had not even heard of LateX before finding this site.
Oh never mind this, it's fixed.
 
Last edited:
  • #5
SammyS
Staff Emeritus
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I don't have much experience with partial derivatives, am I on the right track here: the first partial derivative with respect to x, [tex]\displaystyle\frac{\partial}{\partial x} f(x-ct)[/tex] be: 1, therefore the second would be 0?

No.

[itex]\displaystyle \frac{\partial}{\partial x} f(x-ct) = (1)\cdot f'(x-ct)\,,[/itex] because d/dx (x) = 1 .

Similarly [itex]\displaystyle \frac{\partial}{\partial t} f(x-ct) = (-c)\cdot f'(x-ct)\,.[/itex]
 
  • #6
35
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Oh, so that's what the reference to chain rule was about! Thank you!! :smile:
 

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