Urgent: Differential equation.

**math_Newbie** · February 6th 2008, 12:35 PM

Can anyone give me any pointers with this differential equation.

I seriously have no clue what I am doing with it (More detail the better

)

Find the particular solution of:
x^3 dy/dx = y^3 - 3yx^2 , y(1) = 1

..... remember, I am really unsure what I am doing with this.

Thankyou. math_Newbie.

**TwistedOne151** · February 6th 2008, 01:11 PM

Define $u(x)=x^3y(x)$ . Then $\frac{du}{dx}=x^3\frac{dy}{dx}+3x^2y(x)$
Substituting in our original differential equation:
$\frac{du}{dx}=y^3-3x^2y+3x^2y$
$\frac{du}{dx}=y^3$
$\frac{du}{dx}=\left(\frac{u(x)}{x^3}\right)^3$
$\frac{du}{dx}=\frac{u^3}{x^9}$
Which is a separable equation; after solving for u(x) (remember the constant of integration), you can use $y(x)=\frac{u(x)}{x^3}$ to obtain y(x), and then use y(1)=1 to find the value of the constant.

--Kevin C.

**mr fantastic** · February 6th 2008, 05:04 PM

Originally Posted by math_Newbie

Can anyone give me any pointers with this differential equation.

I seriously have no clue what I am doing with it (More detail the better

)

Find the particular solution of:
x^3 dy/dx = y^3 - 3yx^2 , y(1) = 1

..... remember, I am really unsure what I am doing with this.

Thankyou. math_Newbie.

Alternatively: The DE can be written as $\frac{dy}{dx} = \left( \frac{y}{x} \right)^3 - 3 \left( \frac{y}{x} \right)$ .

For DE's of the form $\frac{dy}{dx} = f\left( \frac{y}{x}\right)$ there's a standard substitution that converts it to first order linear:

$v = \frac{y}{x} \Rightarrow y = xv \,$ and $\, \frac{dy}{dx} = v + \frac{dv}{dx}$ .

If you make this substitution in your case, you get $v + \frac{dv}{dx} = v^3 - 3v \Rightarrow \frac{dv}{dx} = v^3 - 4v$ which is seperable:

$\int dx = \int \frac{dv}{v^3 - 4v}$ .

Since $\frac{1}{v^3 - 4v} = \frac{1}{v(v^2 - 4)} = \frac{1}{v(v - 2)(v + 2)}$ , you have a case for partial fraction decomposition.

Alternatively, and just to stop Krizalid getting an ulcer, in this one instance I will point out that instead of using partial fractions, you might note that

$\frac{1}{v(v^2 - 4)} = \frac{1}{4} \left( \frac{v^2 - (v^2 - 4)}{v(v^2 - 4)} \right) = \frac{1}{4} \left( \frac{v}{v^2 - 4} - \frac{1}{v}\right)$ and "it follows easily"

Either way, once you've solved for v, you've got y: y = xv.

**Krizalid** · February 6th 2008, 05:34 PM

Originally Posted by mr fantastic

Alternatively, and just to stop Krizalid getting an ulcer, in this one instance I will point out that instead of using partial fractions

Amén!

(Just by sayin' that I recently got an ulcer.)

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