differential equations

emily28 · #1 $Old$ July 9th, 2007, 10:59 AM

hey all!

ive got two differentail equations which i really cannot do, but my friend suggested using this website so here goes!

a) xy(dy/dx) - 2(y+3)=0

b) (dy/dx) - y=xy5 (i think its a special one but i dont know how to do it!)

sorry about the writing style, i dont know how to use the special maths tool yet but i hope you understand what i mean :P

thank you

ThePerfectHacker · #2 $Old$ July 9th, 2007, 11:15 AM

Quote:

Originally Posted by emily28 $View Post$

a) [i]xy(dy/dx) - 2(y+3)=0

$xyy' = 2(y+3)$
$\frac{yy'}{2(y+3)} = \frac{1}{x}$ (or $y=-3$ is solution).
THis is seperable.

Krizalid · #3 $Old$ July 9th, 2007, 12:27 PM

Quote:

Originally Posted by emily28 $View Post$

hey all!

b) (dy/dx) - y=xy5

$\begin{aligned}y' - y &= 5xy\\y' &= y(5x + 1)\\\frac{1}{y}~dy &= (5x + 1)~dx\end{aligned}$

Now it's easier.

Krizalid · #4 $Old$ July 9th, 2007, 12:33 PM

$y'-y=xy^5\iff\frac{y'}{y^5}-\frac1{y^4}=x$

If this is the equation, then set $u=\frac1{y^4}$

emily28 · #5 $Old$ July 9th, 2007, 12:52 PM

oh right, but im still unsure where to go from here becuase its seperable but im still slightly confused

Krizalid · #6 $Old$ July 9th, 2007, 12:55 PM

But you didn't tell me what's the right equation

emily28 · #7 $Old$ July 9th, 2007, 12:59 PM

oh right sorry it was

y'-y=xy to the power 5

does this help!!

Krizalid · #8 $Old$ July 9th, 2007, 01:01 PM

Quote:

Originally Posted by Krizalid $View Post$

$y'-y=xy^5\iff\frac{y'}{y^5}-\frac1{y^4}=x$

If this is the equation, then set $u=\frac1{y^4}$

Follow this, 'cause it's a Bernoulli's differential equation.

emily28 · #9 $Old$ July 10th, 2007, 01:03 PM

i understand it upto this point and know its bernolli equation, but i dont know how to solve them at all! could anyone suggest where i should go next

topsquark · #10 $Old$ July 10th, 2007, 08:05 PM

Quote:

Originally Posted by emily28 $View Post$

oh right sorry it was

y'-y=xy to the power 5

does this help!!

Quote:

Originally Posted by Krizalid $View Post$

Follow this, 'cause it's a Bernoulli's differential equation.

Quote:

Originally Posted by emily28 $View Post$

i understand it upto this point and know its bernolli equation, but i dont know how to solve them at all! could anyone suggest where i should go next

Let $u = y^{-4}$ (ie. $y = u^{-1/4}$ .)

Then $\frac{du}{dx} = -4 \cdot y^{-5} \frac{dy}{dx}$

Turning this around we get:
$\frac{dy}{dx} = - \frac{1}{4}y^5 \frac{du}{dx} = - \frac{1}{4} u^{-5/4} \frac{du}{dx}$

So the differential equation becomes:
$y'-y=xy^5$

$- \frac{1}{4} u^{-5/4} \frac{du}{dx} - u^{-1/4} = x \cdot u^{-5/4}$ <-- Multiply both sides by $u^{5/4}$ .

$- \frac{1}{4} \frac{du}{dx} - u = x$

$\frac{du}{dx} + 4 u = -4x$

Now solve this for u(x) and when you get that find y(x).

-Dan

Jhevon · #11 $Old$ July 11th, 2007, 12:39 AM

Quote:

Originally Posted by topsquark $View Post$

$\frac{du}{dx} + 4 u = -4x$

Now solve this for u(x) and when you get that find y(x).

-Dan

i did the manipulation slightly different from what you did, but we ended up with the same thing. i used the modified equation that Krizalid developed to plug the substitutions into. i think it works out nicer that way, but what do i know, maybe it's just nicer in my weird mind

emily28 · #12 $Old$ July 11th, 2007, 07:20 AM

hey, sorry guys im still slightly stumped in how to do the next part...

Jhevon · #13 $Old$ July 11th, 2007, 07:27 AM

Quote:

Originally Posted by emily28 $View Post$

hey, sorry guys im still slightly stumped in how to do the next part...

it's a first order differential equation. multiply through by the integrating factor and simplify. (You know how to find integrating factors right? ...do you know what an integrating factor is?)

topsquark · #14 $Old$ July 11th, 2007, 08:05 AM

Quote:

Originally Posted by topsquark $View Post$

$\frac{du}{dx} + 4 u = -4x$

Quote:

Originally Posted by emily28 $View Post$

hey, sorry guys im still slightly stumped in how to do the next part...

Solve the homogeneous equation first:
$\frac{du_h}{dx} + 4 u_h = 0$

$\frac{du_h}{dx} = -4 u_h$

What function could $u_h(x)$ be? Since we know that $\frac{d}{dx}e^{bx} = be^{x}$ the function $u_h(x)$ must be an exponential. Thus
$u_h(x) = Ae^{bx}$

Plugging this into the homogeneous equation gives:
$Abe^{bx} = -4Ae^{bx}$

This leaves us with b = -4. So $u_h(x) = Ae^{-4x}$ .

Now for the particular solution of $\frac{du}{dx} + 4 u = -4x$ . The RHS is a polynomial so we would guess that the particular solution $u_p(x)$ will be a polynomial. So try $u_p(x) = cx + d$ . Plugging this into the differential equation gives:
$c + 4 (cx + d) = -4x$

$4cx + (c + 4d) = -4x$

So we have the simultaneous equations
$4c = -4$
and
$c + 4d = 0$

Thus c = -1 and d = 1/4. Thus $u_p(x) = -x + \frac{1}{4}$ .

Since the differential equation is linear the most general solution to the differential equation will be the sum of the homogenous and particular solutions. Thus
$u(x) = u_h(x) + u_p(x) = Ae^{-4x} - x + \frac{1}{4}$

Now, to get y(x) we know that $y(x) = -\frac{1}{u^{1/4}}$ . Thus we finally have that
$y(x) = -\frac{1}{(e^{-4x} - x + \frac{1}{4})^{1/4}} = -\frac{1}{\sqrt[4]{e^{-4x} - x + \frac{1}{4}}}$

-Dan

emily28 · #15 $Old$ July 11th, 2007, 08:59 AM

i dont that i was meant to look at particular solutions and expenituals, is there another way to do it using perhaps a different method, as i havent covered this in class and dont think the teacher would want me to approach it this way?