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Q: Solve the separable differential equation:

$3x - 2y\sqrt{x^2+1}\frac{dy}{dx} = 0$ and subject to the initial condition: $y(0) = 4$. What does $y$ equal, in relation to $x$? I am just really confused what I am suppose to do? Thanks.

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This link should help (as a start): – t.b. Mar 30 '11 at 21:58
Thanks, looking into it right now. – Mr_CryptoPrime Mar 30 '11 at 22:04
You'll want to subtract the 3x, divide by the square root, and multiply by dx. Then integrate both sides. This is known as the orangoutang method, but it works ;) – The Chaz 2.0 Mar 30 '11 at 22:08
@The Chaz: heh. orangoutang method, never heard that :). This Wikipedia page is worse written than I feared. In brief the recipe is: all $x$'s on one side, all $y$'s on the other and integrate: Write it as $\int_{0}^{t} \frac{3x}{\sqrt{x^2 + 1}}\,dx = \int_{y(0)}^{y(t)} 2y \,dy$. After finishing the integrals let $t = x$. – t.b. Mar 30 '11 at 22:11
Ok, I think I should be able to do it now! I thought they wanted some partial differentiation crap or something, getting way ahead of Or is this exactly what this is and I am just being an idiot? – Mr_CryptoPrime Mar 30 '11 at 22:16
up vote 2 down vote accepted

So what we want here is a particular solution to our ODE given our initial condition: $y(0) = 4$

$$3x\ dx = 2y\sqrt{x^{2}+1}\ dy~;\hspace{20 pt} y(0)=4$$

$$\Rightarrow \displaystyle\int 2y\ dy = \displaystyle\int \dfrac{3x}{\sqrt{x^{2}+1}}\ dx$$

$u = x^{2}+1$

$du = 2x\ dx$

$dx = \dfrac{1}{2x}du$

Making the substitutions in for u and dx we see we come to the following:

$~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~\Rightarrow y^{2} = 3 \cdot \dfrac{1}{2} \displaystyle\int \dfrac{x}{\sqrt{u}} \cdot \dfrac{1}{x}\ du$

$~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~\Rightarrow y^{2} = \dfrac{3}{2} \displaystyle\int \dfrac{1}{\sqrt{u}}\ du$

$~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~\Rightarrow y^{2} = \dfrac{3}{2} \displaystyle\int u^{-\tfrac{1}{2}}\ du$

$~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~\Rightarrow y^{2} = \dfrac{3}{2} \cdot 2 ~ u^{\tfrac{1}{2}} + C$

$~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~\Rightarrow y^{2} = 3 \sqrt{u}\ du$

$~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~\Rightarrow y^{2} = 3 \cdot \sqrt{x^{2}+1} + C$

$~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~\Rightarrow y(x) = \displaystyle\pm \sqrt{3} \sqrt[4]{x^{2}+1} + C$

$\underline{\text{Applying Initial Conditions:}}$

$y(0)=4:~~4=\sqrt{3} \displaystyle\sqrt[4]{(0)^{2}+1}~ + C$

$~~~~~~~~~~~~~~~~~~~4=\sqrt{3}\cdot 1~ + C$


$y(0)=4:~~4=-\sqrt{3} \displaystyle\sqrt[4]{(0)^{2}+1}~ + C$

$~~~~~~~~~~~~~~~~~~~4=-\sqrt{3}\cdot 1~ + C$


Thus the particular solutions to this separable ODE is,

$~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~\therefore~~y(x) = \displaystyle \sqrt{3} \sqrt[4]{x^{2}+1} + 2.3 ~~~~\&~~~~ y(x) = \displaystyle -\sqrt{3} \sqrt[4]{x^{2}+1} + 5.7~~~~\blacksquare$

I hope this helped out, and hopefully I did not make any mistakes to cause any type of confusion.

Let me know if there is any step that did not make since to you and I will try and clarify a some more.


Good Luck.

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Here's what you are supposed to do

$$\int \frac{3x}{\sqrt{x^2+1}}dx = \int 2y dy + C$$

You have a given initial condition and an undetermined constant..

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[\begin{array}{l} 3x - 2y\sqrt {{x^2} + 1} \frac{{{\rm{d}}y}}{{{\rm{d}}x}} = 0\\ 3x = 2y\sqrt {{x^2} + 1} \frac{{{\rm{d}}y}}{{{\rm{d}}x}}\\ 3\frac{x}{{\sqrt {{x^2} + 1} }} = 2y\frac{{{\rm{d}}y}}{{{\rm{d}}x}}\\ \frac{3}{2}\int_{}^{} {\frac{x}{{\sqrt {{x^2} + 1} }}{\rm{d}}x} = \int_{}^{} {y{\rm{ d}}y} \\ \frac{3}{2}\int_{}^{} {\tan \theta \sec \theta {\rm{ d}}\theta } = \frac{{{y^2}}}{2} + {C_1}\\ \frac{3}{2}\sec \theta + \frac{3}{2}{C_2} = \frac{{{y^2}}}{2} + {C_1}\\ 3\sec \arctan x + C = {y^2}\\ {y^2} = 3\sqrt {1 + {x^2}} + C\\ y = \left( {{\mathop{\rm sgn}} k} \right)\sqrt {3\sqrt {1 + {x^2}} + C} \\ 4 = y(0) = \pm \sqrt {3 + C} \\ 4 = \sqrt {3 + C} \\ C = 13\\ {\mathop{\rm sgn}} k = 1\\ y = \sqrt {3\sqrt {1 + {x^2}} + 13} \end{array}]

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