Thread: [SOLVED] Few ODEs I can't solve

1. [SOLVED] Few ODEs I can't solve

1st.

I solved the following question :-

$(\sqrt{x+y} + \sqrt{x-y} )dx - (\sqrt{x+y} - \sqrt{x-y} )dy = 0$

I could solve it till the following step. Got stuck after that.

$\frac{2Vx - v( \sqrt {x+Vx} - \sqrt {x-Vx} ) } { (\sqrt {x+Vx} - \sqrt {x-Vx} ) }= \frac{xdv}{dx}$

Is there any mistake in this? Please solve it further.

2. Originally Posted by Altair
1st.

I solved the following question :-

$(\sqrt{x+y} + \sqrt{x-y} )dx - (\sqrt{x+y} - \sqrt{x-y} )dy = 0$

I could solve it till the following step. Got stuck after that.

$\frac{2Vx - v( \sqrt {x+Vx} - \sqrt {x-Vx} ) } { (\sqrt {x+Vx} - \sqrt {x-Vx} ) }= \frac{xdv}{dx}$

Is there any mistake in this? Please solve it further.
$\frac{dy}{dx} = \frac{\sqrt{x + y} + \sqrt{x - y}}{\sqrt{x + y} - \sqrt{x - y}} = \frac{\sqrt{x} \, \sqrt{1 + \frac{y}{x}} + \sqrt{x} \, \sqrt{1 - \frac{y}{x}}}{\sqrt{x} \, \sqrt{1 + \frac{y}{x}} - \sqrt{x} \, \sqrt{1 - \frac{y}{x}}} = \frac{\sqrt{1 + \frac{y}{x}} + \sqrt{1 - \frac{y}{x}}}{\sqrt{1 + \frac{y}{x}} - \sqrt{1 - \frac{y}{x}}}$

$\Rightarrow v + x \, \frac{dv}{dx} = \frac{\sqrt{1 + v} + \sqrt{1 - v}}{\sqrt{1 + v} - \sqrt{1 - v}} = \frac{(\sqrt{1 + v} + \sqrt{1 - v})}{(\sqrt{1 + v} - \sqrt{1 - v})} \times \frac{(\sqrt{1 + v} + \sqrt{1 - v})}{(\sqrt{1 + v} + \sqrt{1 - v})}$

$\frac{1 + \sqrt{1 - v^2}}{v}$

(after expanding and simplifying).

Therefore $x \frac{dv}{dx} = \frac{1 + \sqrt{1 - v^2}}{v} - v = \frac{1 - v^2 + \sqrt{1 - v^2}}{v}$.

Therefore $\int \frac{v}{1 - v^2 + \sqrt{1 - v^2}} \, dv = \int \frac{1}{x} \, dx$.

I'm afraid I might have to sleep on it for a while (since it's after midnight right now for me) as I don't see any obvious next step right now.

I'll make a further reply later if no-one else has stepped in.

3. Originally Posted by mr fantastic
$\frac{dy}{dx} = \frac{\sqrt{x + y} + \sqrt{x - y}}{\sqrt{x + y} - \sqrt{x - y}} = \frac{\sqrt{x} \, \sqrt{1 + \frac{y}{x}} + \sqrt{x} \, \sqrt{1 - \frac{y}{x}}}{\sqrt{x} \, \sqrt{1 + \frac{y}{x}} - \sqrt{x} \, \sqrt{1 - \frac{y}{x}}} = \frac{\sqrt{1 + \frac{y}{x}} + \sqrt{1 - \frac{y}{x}}}{\sqrt{1 + \frac{y}{x}} - \sqrt{1 - \frac{y}{x}}}$

$\Rightarrow v + x \, \frac{dv}{dx} = \frac{\sqrt{1 + v} + \sqrt{1 - v}}{\sqrt{1 + v} - \sqrt{1 - v}} = \frac{(\sqrt{1 + v} + \sqrt{1 - v})}{(\sqrt{1 + v} - \sqrt{1 - v})} \times \frac{(\sqrt{1 + v} + \sqrt{1 - v})}{(\sqrt{1 + v} + \sqrt{1 - v})}$

$\frac{1 + \sqrt{1 - v^2}}{v}$

(after expanding and simplifying).

Therefore $x \frac{dv}{dx} = \frac{1 + \sqrt{1 - v^2}}{v} - v = \frac{1 - v^2 + \sqrt{1 - v^2}}{v}$.

Therefore $\int \frac{v}{1 - v^2 + \sqrt{1 - v^2}} \, dv = \int \frac{1}{x} \, dx$.

I'm afraid I might have to sleep on it for a while (since it's after midnight right now for me) as I don't see any obvious next step right now.

I'll make a further reply later if no-one else has stepped in.
The integral on the left is equal to $-\ln |\sqrt{1 - v^2} + 1|$.

Proof posted later unless someone steps in first.

4. Originally Posted by mr fantastic
The integral on the left is equal to $-\ln |\sqrt{1 - v^2} + 1|$.

Proof posted later unless someone steps in first.
Make the substitution $u = \sqrt{1 - v^2}$.

Then

$\frac{du}{dv} = \frac{-v}{\sqrt{1 - v^2}} = - \frac{v}{u} \Rightarrow dv = - \frac{u}{v} \, du$.

Then the integral becomes

$\int \frac{v}{u^2 + u} \, \left( - \frac{u}{v} \right) = - \int \frac{u}{u^2 + u} \, du = - \int \frac{1}{u + 1} \, du, \, u \neq 0 \,$,

$= - \ln |u + 1| + C = - \ln |\sqrt {1 - v^2} + 1| + C$.

5. Originally Posted by mr fantastic
$\frac{dy}{dx} = \frac{\sqrt{x + y} + \sqrt{x - y}}{\sqrt{x + y} - \sqrt{x - y}} = \frac{\sqrt{x} \, \sqrt{1 + \frac{y}{x}} + \sqrt{x} \, \sqrt{1 - \frac{y}{x}}}{\sqrt{x} \, \sqrt{1 + \frac{y}{x}} - \sqrt{x} \, \sqrt{1 - \frac{y}{x}}} = \frac{\sqrt{1 + \frac{y}{x}} + \sqrt{1 - \frac{y}{x}}}{\sqrt{1 + \frac{y}{x}} - \sqrt{1 - \frac{y}{x}}}$

$\Rightarrow v + x \, \frac{dv}{dx} = \frac{\sqrt{1 + v} + \sqrt{1 - v}}{\sqrt{1 + v} - \sqrt{1 - v}} = \frac{(\sqrt{1 + v} + \sqrt{1 - v})}{(\sqrt{1 + v} - \sqrt{1 - v})} \times \frac{(\sqrt{1 + v} + \sqrt{1 - v})}{(\sqrt{1 + v} + \sqrt{1 - v})}$

$\frac{1 + \sqrt{1 - v^2}}{v}$

(after expanding and simplifying).

Therefore $x \frac{dv}{dx} = \frac{1 + \sqrt{1 - v^2}}{v} - v = \frac{1 - v^2 + \sqrt{1 - v^2}}{v}$.

Therefore $\int \frac{v}{1 - v^2 + \sqrt{1 - v^2}} \, dv = \int \frac{1}{x} \, dx$
Continuing:

$- \ln |\sqrt{1 - v^2} + 1| = \ln |x| + C$

$\Rightarrow \ln \left| \frac{1}{\sqrt{1 - v^2} + 1} \right| = \ln |x| + C$

$\Rightarrow \frac{1}{\sqrt{1 - v^2} + 1} = Ax \,$ where $A = e^C$

$\Rightarrow \sqrt{1 - v^2} + 1 = \frac{B}{x},$ where $B = \frac{1}{A}$

$\Rightarrow 1 - v^2 = \left( \frac{B}{x} - 1\right)^2$.

Substitute from $y = xv$ and you can probably see what the final couple of lines leading to an answer will be .....