First order DE

  • Thread starter fluidistic
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  • #1
fluidistic
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Homework Statement


Find the general and singular solutions of the following DE:
[itex]x^2 (y')^2-2(xy-4)y'+y^2=0[/itex].


Homework Equations


Not sure. A friend of mine told me it's a Clairaut's equation but I couldn't even reduce the given one into a Clairaut's equation.
[itex]y(x)=xy'+f(y')[/itex]


The Attempt at a Solution


Literally stuck on starting. I checked out for the homogeneity of the DE but it's not. Maybe Bernoulli can come into action... but I'm not sure.
Any tip is greatly appreciated, as usual.
 

Answers and Replies

  • #2
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i aint no wizz kid.... and ive never seen clairauts before... but...

take sq root of both sides and rearrange for your general form?
 
  • #3
fluidistic
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i aint no wizz kid.... and ive never seen clairauts before... but...

take sq root of both sides and rearrange for your general form?

I thought about that but there's an "y" term in front of the y' term, so I can't isolate y(x) if I simply take the square root.
 
  • #4
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cant isolate it completely but could it just be an implicit solution? where f is a function of x and y?
like i said i dunno about clairauts method so ill stop spamming ya thread haha
 
  • #5
fluidistic
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No this isn't an implicit solution I think. I'd get y in function of y, y', y'' and x which isn't a success. :)
I'll wait for further help.
 
  • #6
vela
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(xy'-y)2 is pretty close to the LHS. Does that help?
 
  • #7
fluidistic
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(xy'-y)2 is pretty close to the LHS. Does that help?

Ahahaha, I just did this on my draft. The DE reduces to [itex](xy'-y)^2+8y'[/itex].
I'll think on how to continue.
 
  • #8
fluidistic
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The farest expression I reach is [itex]y=xy' \mp 2 \sqrt 2 (-y')^{1/2}[/itex]. I would have thought to factorize by y' and then separate variables but the y's don't have the same exponent number so this doesn't seem to work.
I'm once again stuck.
 
  • #9
vela
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Now you want to use the standard approach to solving Clairaut's equation. Is that what you tried?
 
  • #10
fluidistic
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Now you want to use the standard approach to solving Clairaut's equation. Is that what you tried?
No, I "forgot" about Clairaut's equation, I thought my friend made a mistake.
So thank you for pointing this out.
I reach [itex]y(x)=c_1x+c_2[/itex]. To verify if this work, I plugged it into the original DE and I got a condition on the constant for the DE to be satisfied. Namely [itex]c_1[/itex] must be equal to [itex]-\frac{c_2 ^2}{8}[/itex].
So the final solution can be written on the form [itex]y(x)=\frac{-c^2}{8}x+c^2[/itex].
Is it possible that I get such a condition for the constants without having been given initial conditions? (I think yes and that the initial conditions will determine c, but I am not 100% sure)
 
  • #11
vela
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It's a first-order equation so you only want one constant left at the end to be determined by the initial condition. The other one should be adjusted to satisfy the particular differential equation, just like you found.
 
  • #12
fluidistic
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Okay, thank you vela. :smile:
 

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