1. differential equations

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

2. Originally Posted by emily28
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.

3. Originally Posted by emily28
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.

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

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

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

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

7. oh right sorry it was

y'-y=xy to the power 5

does this help!!

8. Originally Posted by Krizalid
$y'-y=xy^5\iff\frac{y'}{y^5}-\frac1{y^4}=x$

If this is the equation, then set $u=\frac1{y^4}$
Follow this, 'cause it's a Bernoulli's differential equation.

9. 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

10. Originally Posted by emily28
oh right sorry it was

y'-y=xy to the power 5

does this help!!
Originally Posted by Krizalid
Follow this, 'cause it's a Bernoulli's differential equation.
Originally Posted by emily28
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

11. Originally Posted by topsquark

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

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

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

13. Originally Posted by emily28
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?)

14. Originally Posted by topsquark
$\frac{du}{dx} + 4 u = -4x$
Originally Posted by emily28
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

15. 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?

Page 1 of 2 12 Last