Differential Equation Substitution

  • Thread starter KillerZ
  • Start date
  • #1
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Homework Statement



Solve the given differential equation by using an appropriate substitution.

Homework Equations



[tex]x\frac{dy}{dx} = y + \sqrt{x^{2} - y^{2}}[/tex], x > 0

The Attempt at a Solution



[tex]x\frac{dy}{dx} = y + \sqrt{x^{2} - y^{2}}[/tex]

[tex]xdy = (y + \sqrt{x^{2} - y^{2}})dx[/tex]

[tex]y = ux[/tex]

[tex]u = \frac{y}{x}[/tex]

[tex]dy = udx + xdu[/tex]

[tex]x[udx + xdu] = (ux + \sqrt{x^{2} - u^{2}x^{2}})dx[/tex]

[tex]xudx + x^{2}du = uxdx + x\sqrt{1 - u^{2}}dx[/tex]

[tex]\frac{du}{\sqrt{1 - u^{2}}} = \frac{dx}{x}[/tex]

[tex]\frac{du}{\sqrt{1 - u^{2}}} - \frac{dx}{x} = 0[/tex]

[tex]\int\frac{du}{\sqrt{1 - u^{2}}} - \int\frac{dx}{x} = 0[/tex]

[tex]sin^{-1}(u) - ln|x| = ln|c|[/tex]

[tex]sin^{-1}(\frac{y}{x}) - ln|x| = ln|c|[/tex]

[tex]e^{sin^{-1}(\frac{y}{x})} - x = c[/tex]
 
Last edited:

Answers and Replies

  • #2
109
0
Hi KillerZ,

Look perfect to me except the very end. You have:

[tex]
arcsin(u) - ln|x| = ln|c|
[/tex]

the next step isn't quite right, it should go:

[tex]
arcsin(u) = ln|x| + ln|c|
[/tex]

[tex]
arcsin\left(\frac{y}{x}\right) = ln|cx|
[/tex]

[tex]
e^{arcsin\left(\frac{y}{x}\right)} = cx
[/tex]

but other than that its perfect, I think you'll probably realise the mistake when you see, just a momentary lapse of judgement. I will say you didn't really need to do this

[tex]
\frac{du}{\sqrt{1 - u^{2}}} - \frac{dx}{x} = 0
[/tex]

in fact its usally best to keep the integration of different variables on different sides, just as you did in you step before this:

[tex]
\int \frac{1}{\sqrt{1 - u^{2}}} \ du \ = \ \int \frac{1}{x} \ dx
[/tex]

I suppose the mistake at the end is a perfect example of why. Have fun KillerZ :D
 
Last edited:
  • #3
rock.freak667
Homework Helper
6,223
31
From this line

[tex]
sin^{-1}(\frac{y}{x}) - ln|x| = ln|c|
[/tex]

if you raise both sides to e then you need to use

ea+b=ea*eb
 
  • #4
116
0
Ok thanks I just messed up my log rules.
 

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