2 differential equations

By Anonymous on Sunday, September 3, 2000 - 02:52 pm :

Hi How would you solve these equations, I am getting different answers and am can't find the error in any of the methods I am trying.

dy/dx=(x+y)/(x-y)

dy/dx + xy = x

By Brad Rodgers (P1930) on Sunday, September 3, 2000 - 11:05 pm :

For the second one, solve for dy/dx,

dy/dx=x-xy
=x(1-y)

Thus, 1/(1-y) x dy=x x dx.

This translates to

\int 1 / (1 - y) \times dy = \int x \times dx

Now,

$\int x \times dx = x^{2} / 2$
and, using the same logic that is found in Integration by substitution , we know that

$\int 1 / (1 - y) \times dy = - \ln (y - 1)$

So,

$x^{2} / 2 = - \ln (y + 1)$

$y = e^{- x^{2}} / 2 + 1$
The first problem actually seems a good deal more difficult, so I really don't know how to answer it.

Brad

By David Loeffler (P865) on Monday, September 4, 2000 - 12:07 am :

Sorry to interfere, but I think I have found an expression for x in the first equation in terms of y/x. I know this isn't a lot of help, but somebody else might be able to finish it off.

Let v=y/x. The RHS
becomes (1+v)/(1-v), and the LHS becomes x(dv/dx)+v.
Take v across to the right:

x(dv/dx)=(1+v² )/(1-v)

Rearranged, this is:

(1-v)/(1+v² ) dv = 1/x dx

This integrates reasonably easily to give

arctan(v)-½ln(1+v² ) = ln x

or x = e^{arctan v} /sqrt(1+v² ).

But I haven't a clue how one would go about finding v in terms of x from this.

David

By David Loeffler (P865) on Monday, September 4, 2000 - 01:16 am :

DOH! The final expression simplifies wonderfully. We have
x sqrt(1+v² )=e^{arctan v}
sqrt(x² +y² )=e^{arctan y/x}

Suppose we change to radial coordinates

r = e^{θ}

!!
So the solution is a logarithmic spiral.

(Actually there is an arbitrary constant in there because of the constant of integration, but that's minor. If you want you can substitute for x and y in terms of theta into the original differential equation - it comes out as a solution.)

Bye now - at this rate I'm going to fall asleep in front of my screen

David

By Tim Garton (P2843) on Monday, September 4, 2000 - 07:11 pm :

Thanks everyone

By David Loeffler (P865) on Monday, September 4, 2000 - 08:45 pm :

By the way, where did you get these equations from in the first place?

David

By Tim Garton (P2843) on Tuesday, September 5, 2000 - 08:19 pm :

Work I was sent to do before I go to university.

By Anonymous on Thursday, September 7, 2000 - 08:32 pm :

In the conversation above, there are the following lines:

\int 1 / (1 - y) dy = - \ln (y - 1)

and with $dy / dx = (x + y) / (x - y)$ when substituting $v = y / x$ the lhs becomes $x (dv / dx) + v$ .
Could someone explain how these work.

By Brad Rodgers (P1930) on Friday, September 8, 2000 - 11:52 pm :

b = \int 1 / (1 - y) dy

Then if $u = y - 1$

$du / dy = 1$ , $du = dy$

and then

$b = \int - 1 / u du = - \ln (u) = - \ln (y - 1)$
That should explain the first one.

Brad

By Anonymous on Saturday, September 9, 2000 - 12:18 pm :

Thank you

By Tony Chi Kin Ho (P1942) on Saturday, September 9, 2000 - 04:30 pm :

What about the second one? I can't figure it out either!

And I have got another one related:

Solve:

dy/dx + 5y = e^8x , given that y = 3/2 when x = 0.

By Pras Pathmanathan (Pp233) on Saturday, September 9, 2000 - 08:04 pm :

OK, for the second one, we want to know what

dy / dx

becomes when we use the substitution

v = y / x

$v = y / x$ , so $y = v x$ .

Differentiate both sides with respect to $x$ , and the lhs becomes $dy / dx$ , while the rhs becomes:

$v (dx / dx) + x (dv / dx)$ (using the product rule)

and $dx / dx = 1$ , so we have $dy / dx = v + x (dv / dx)$ .

As for solving $dy / dx + 5 y = e^{8 x}$ , you need to know about a clever method for solving such differential equations involving integrating factors. They are used on equations of the form

$dy / dx + p (x) y = q (x)$

for any functions of $x$ , $p$ and $q$ . (So in your equation, $p (x) = 5$ , and $q (x) = e^{8 x}$ .)

Now, firstly you work out what $e^{\int p (x) dx}$ is.

In your case, $p (x) = 5$ , so $\int p (x) dx = 5 x$ (don't worry about constants), and so

$e^{\int p (x) dx} = e^{5 x}$

Now the clever bit: multiply both sides of your diff. eqn by this $e^{5 x}$ , so you have

$e^{5 x} (dy / dx) + e^{5 x} 5 y = e^{8 x} e^{5 x} (= e^{13 x})$

and then notice that the lhs is just

$d (e^{5 x} y) / dx$

(or in general, $d (e^{\int p (x) dx} y) / dx$

so you have the eqn:

$d (e^{5 x} y) / dx = e^{13 x}$

which you can solve by integrating both sides (remembering the constant this time) and then solving for $y$ . You then get rid of the constant of integration with your initial condition: $y (x = 0) = 3 / 2$ .

Hope that helps, write back if you want me to explain any part of that in more detail.

Pras