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Diff EQNot sure what method to use

  • #1
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2

Homework Statement



In one of my fluids homeworks, I wound up with a DE of the form

[tex]A\frac{dh}{dt}+B\sqrt{h}+C=0[/tex]

where A, B, and C are known constants. Can I use the integrating factor on this if I divide thorough by the
leading coefficient A and put it in the form

[tex]\frac{dh}{dt}+\frac{B}{A}\sqrt{h}=-\frac{C}{A}[/tex]

?

I am just not sure how to approach this DE. Any hints are great :smile:
 
Last edited:

Answers and Replies

  • #2
dx
Homework Helper
Gold Member
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18
This is seperable.
 
  • #3
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2
This is seperable.
Okay. Is it separable as is? Because I don't see it? Perhaps a substitution is needed?
 
  • #4
dx
Homework Helper
Gold Member
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Yes, just divide by B√h + C, and multiply by dt. You will get

[tex] \frac{-A}{B\sqrt{h} + C}dh = dt [/tex]
 
  • #5
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2
Wow. I need to brush up on my math, because I still have no idea how you did that. I have no idea how to divide

[tex]
A\frac{dh}{dt}+B\sqrt{h}+C=0
[/tex]

by B√h + C

I wouldn't even know where to start. I am trying now. Do you do it term by term? I realize that this is just Algebra, but I guess mt Engineering classes
are veering away from all that goodness
 
  • #6
dx
Homework Helper
Gold Member
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18
[tex] A\frac{dh}{dt}+B\sqrt{h}+C=0 [/tex]

[tex] B\sqrt{h}+C = -A\frac{dh}{dt} [/tex]

Dividing by B√h + C,

[tex] 1 = \frac{-A}{B\sqrt{h}+C}\frac{dh}{dt} [/tex]

Multiplying by dt,

[tex] dt = \frac{-A}{B\sqrt{h} + C}dh [/tex]
 
  • #7
3,003
2
[tex] A\frac{dh}{dt}+B\sqrt{h}+C=0 [/tex]

[tex] B\sqrt{h}+C = -A\frac{dh}{dt} [/tex]

Dividing by B√h + C,

[tex] 1 = -A\frac{dh}{dt} \frac{1}{B\sqrt{h}+C} [/tex]

Multiplying by dt,

[tex] dt = \frac{-A}{B\sqrt{h} + C}dh [/tex]
Wow :blushing: I had to see it to believe it! Why can't I see these things?!
I can bang my head all effin' day trying to figure this stuff out and then wham! Some
one comes along and is like "hey, why don't you just do this?" and in 2 seconds they're done!

Thanks dx! I am going to go burn something down now :smile:
 

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