Differential Equation problem

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  • #1
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Homework Statement


[itex]x^2y\frac{d^2y}{dx^2} + (x\frac{dy}{dx} - y)^2 = 0[/itex]

[itex]y^2=C_{1}x^2+C_{2}x[/itex]

[itex]y=2;y'=1 when x=1[/itex]

Homework Equations


Can someone give me the answer? Just the answer, the particular solution. Please? I'm having really hard time calculating for it. This is one of the seatworks given to us, and this is the remaining problem that I cannot answer.

The Attempt at a Solution


differentiating the second equation
[itex]y'=\frac{2C_{1}x+C_{2}}{2y}[/itex]
[itex]y''=\frac{4C_{1}y-2C_{1}^2x^2+C_{2}^2+3C_{1}C_{2}x}{4y^2}[/itex]
I'm stuck right there. I substituted the y' and y'' to the first equation but the answer seems not equal to zero. Is my differentiation right?
 
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Answers and Replies

  • #2
vela
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The numerator of the first derivative isn't correct. Perhaps a typo? The second derivative looks wrong too.

Is the point of the problem to just find the constants c1 and c2? If so, you only need to calculate the first derivative.
 
  • #3
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I see, I forgot to place the x right after the C1? Is that what you're talking about? :biggrin: Fixed for you. :smile:

Why first derivative only? The problem (first equation) requires for the 2nd derivative, which makes it really hard (In my part).
 
  • #4
vela
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If you're only solving for the constants, you do that using the two equations y'(1)=1 and y(1)=2. You only need the expressions for y(x) and y'(x) to write down those two equations.

If you're also supposed to verify that y(x) is indeed a solution to the differential equation, you will need y''(x).
 
  • #5
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I see. I substituted the values y'=1; y=2 when x=1, and this is what I got.

[itex]C_{1}=0[/itex]

[itex]C_{2}=4[/itex]

Then, the particular solution should be:
[itex]y=2\sqrt{x}[/itex]

Am I correct?
 
  • #6
vela
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Yes, that function satisfies the differential equation and the initial conditions.
 
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