# Differential equations

• Feb 14th 2008, 06:30 AM
dankelly07
Differential equations
I'm after some help with these two equations.

My problem is factorising with equations like this, I can never tell when I'm supposed to do it!! I have the answers to these questions I just need explicit help, basically a walk through of these examples, if anyone can help I'd much appreciate it,
Thanks

ydy/dx+7x^2+7x+5=0

which satisfies y(0)=4

sqrt(16-(14/3x^3 + 7x^2+10x))

and my other problem is,

dy/dx = 1+y^2/7+9x

which satisfies y(1) = 0

tan(1/9ln(|7+9x/16|))

With regards to the last answer, how does PI sometimes come out of this answer?

PS how do you write with math symbols, it looks a bit messy....
• Feb 14th 2008, 06:34 AM
Krizalid
Both of them are separable.

(See my signature for LaTeX typesetting.)
• Feb 14th 2008, 07:58 AM
wingless
A separable differential equation is an equation that can be expressed in this form:
$\displaystyle \frac{dy}{dx} = g(x)\cdot h(y)$

Like, $\displaystyle \frac{dy}{dx} = 4xy^2,~ \frac{dy}{dx} = \frac{x^2y-y}{x+1},~ \frac{dy}{dx} = \frac{xy^3}{\sqrt{x^2+1}}~.~.~.$

You can solve them by separating the functions and differentials to the sides and integrating.

For example,

$\displaystyle \frac{dy}{dx} = 2xy^2$

Move $\displaystyle dx$ to the right and $\displaystyle y^2$ to the left.

$\displaystyle \frac{1}{y^2}~dy = 2x~dx$

Now you can integrate both sides.

$\displaystyle \int \frac{1}{y^2}~dy = \int 2x~dx$

$\displaystyle -\frac{1}{y} + C_1 = x^2 + C_2$

We can use only C instead of two integration constants.
$\displaystyle -\frac{1}{y} = x^2 + C$

$\displaystyle y = - \frac{1}{x^2 + C}$

-------------------------------------

$\displaystyle y\frac{dy}{dx} +7x^2+7x+5=0$, $\displaystyle y(0)=4$

Start by separating.

$\displaystyle y~dy = -7x^2-7x-5~dx$

Now we can integrate both sides.

$\displaystyle \int y~dy = \int -7x^2-7x-5~dx$

$\displaystyle \frac{y^2}{2}+C_1 = -\frac{7}{3}x^3 -\frac{7}{2}x^2-5x +C_2$

$\displaystyle \frac{y^2}{2} = -\frac{7}{3}x^3 -\frac{7}{2}x^2-5x +C$

$\displaystyle y^2 = -\frac{14}{3}x^3 -7x^2-10x +2C$

$\displaystyle |y| = \sqrt{-\frac{14}{3}x^3 -7x^2-10x +2C}$

-----$\displaystyle y = \sqrt{-\frac{14}{3}x^3 -7x^2-10x +2C}$

-----$\displaystyle y = -\sqrt{-\frac{14}{3}x^3 -7x^2-10x +2C}$

Try $\displaystyle y(0) = 4$ for both of them.

-----$\displaystyle y = \sqrt{-\frac{14}{3}x^3 -7x^2-10x +2C}$
-----$\displaystyle 4 = \sqrt{2C}$
-----$\displaystyle \boxed{2C = 16}$

-----$\displaystyle y = -\sqrt{-\frac{14}{3}x^3 -7x^2-10x +2C}$
-----$\displaystyle 4 = -\sqrt{2C}$
-----As $\displaystyle \sqrt{2C}$ is always positive, this equation doesn't have a root.

So we found that our function is $\displaystyle y = \sqrt{-\frac{14}{3}x^3 -7x^2-10x +2C}$ because this is the only valid function that makes $\displaystyle y(0) = 4$. And we found that $\displaystyle 2C = 16$.

Hence, the function we are looking for is:
$\displaystyle y = \sqrt{-\frac{14}{3}x^3 -7x^2-10x + 16}$

$\displaystyle y = \sqrt{16-(\frac{14}{3}x^3 +7x^2+10x)}$

-------------------------------------

Now try to solve the second equation (Wink)