
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
The Answer is:
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....

Both of them are separable.
(See my signature for LaTeX typesetting.)

A separable differential equation is an equation that can be expressed in this form:
$\displaystyle \frac{dy}{dx} = g(x)\cdot h(y)$
Like, $\displaystyle \frac{dy}{dx} = 4xy^2,~ \frac{dy}{dx} = \frac{x^2yy}{x+1},~ \frac{dy}{dx} = \frac{xy^3}{\sqrt{x^2+1}}~.~.~.$
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.
$\displaystyle \frac{1}{y} = x^2 + C$
$\displaystyle y =  \frac{1}{x^2 + C}$

Your first question is,
$\displaystyle y\frac{dy}{dx} +7x^2+7x+5=0$, $\displaystyle y(0)=4$
Start by separating.
$\displaystyle y~dy = 7x^27x5~dx$
Now we can integrate both sides.
$\displaystyle \int y~dy = \int 7x^27x5~dx$
$\displaystyle \frac{y^2}{2}+C_1 = \frac{7}{3}x^3 \frac{7}{2}x^25x +C_2$
$\displaystyle \frac{y^2}{2} = \frac{7}{3}x^3 \frac{7}{2}x^25x +C$
$\displaystyle y^2 = \frac{14}{3}x^3 7x^210x +2C$
$\displaystyle y = \sqrt{\frac{14}{3}x^3 7x^210x +2C}$
$\displaystyle y = \sqrt{\frac{14}{3}x^3 7x^210x +2C}$
$\displaystyle y = \sqrt{\frac{14}{3}x^3 7x^210x +2C}$
Try $\displaystyle y(0) = 4$ for both of them.
$\displaystyle y = \sqrt{\frac{14}{3}x^3 7x^210x +2C}$
$\displaystyle 4 = \sqrt{2C}$
$\displaystyle \boxed{2C = 16}$
$\displaystyle y = \sqrt{\frac{14}{3}x^3 7x^210x +2C}$
$\displaystyle 4 = \sqrt{2C}$
As $\displaystyle \sqrt{2C}$ is always positive, this equation doesn't have a root.
So we found that our function is $\displaystyle y = \sqrt{\frac{14}{3}x^3 7x^210x +2C}$ 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:
$\displaystyle y = \sqrt{\frac{14}{3}x^3 7x^210x + 16}$
$\displaystyle y = \sqrt{16(\frac{14}{3}x^3 +7x^2+10x)}$

Now try to solve the second equation (Wink)