Differential Equation Substitution

[KillerZ]

Homework Statement

Solve the given differential equation by using an appropriate substitution.

Homework Equations

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

The Attempt at a Solution

x\frac{dy}{dx} = y + \sqrt{x^{2} - y^{2}}

xdy = (y + \sqrt{x^{2} - y^{2}})dx

y = ux

u = \frac{y}{x}

dy = udx + xdu

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

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

\frac{du}{\sqrt{1 - u^{2}}} = \frac{dx}{x}

\frac{du}{\sqrt{1 - u^{2}}} - \frac{dx}{x} = 0

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

sin^{-1}(u) - ln|x| = ln|c|

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

e^{sin^{-1}(\frac{y}{x})} - x = c

 

[Galadirith]

Hi KillerZ,

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

<br /> arcsin(u) - ln|x| = ln|c|<br />

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

<br /> arcsin(u) = ln|x| + ln|c| <br />

<br /> arcsin\left(\frac{y}{x}\right) = ln|cx|<br />

<br /> e^{arcsin\left(\frac{y}{x}\right)} = cx<br />

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

<br /> \frac{du}{\sqrt{1 - u^{2}}} - \frac{dx}{x} = 0<br />

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

<br /> \int \frac{1}{\sqrt{1 - u^{2}}} \ du \ = \ \int \frac{1}{x} \ dx<br />

I suppose the mistake at the end is a perfect example of why. Have fun KillerZ :D

 

[rock.freak667]

From this line

<br /> sin^{-1}(\frac{y}{x}) - ln|x| = ln|c|<br />

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

e^a+b=e^a*e^b

 

[KillerZ]

Ok thanks I just messed up my log rules.