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:

$$arcsin(u) - ln|x| = ln|c|$$

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

$$arcsin(u) = ln|x| + ln|c|$$

$$arcsin\left(\frac{y}{x}\right) = ln|cx|$$

$$e^{arcsin\left(\frac{y}{x}\right)} = cx$$

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

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

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

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

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

 

[rock.freak667]

From this line

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

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.