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


ò1/(1-y)×dy=òx×dx

Now,

òx×dx=x2/2
and, using the same logic that is found in Integration by substitution , we know that

ò1/(1-y)×dy=-ln(y-1)

So,

x2/2=-ln(y+1)

y=e-x2/2+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 = earctan 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 )=earctan v
sqrt(x2 +y2 )=earctan y/x

Suppose we change to radial coordinates


r=eq!!
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:

ò1/(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 :

If


b=ò1/(1-y)dy

Then if u=y-1

du/dy=1, du=dy

and then

b=ò-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 = e8x , 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+5y=e8x, 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)=e8x.)

Now, firstly you work out what eòp(x) dx is.

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

eòp(x) dx=e5x

Now the clever bit: multiply both sides of your diff. eqn by this e5x, so you have

e5x (dy/dx)+e5x 5y=e8xe5x(=e13x)

and then notice that the lhs is just

d(e5xy)/dx

(or in general, d(eòp(x) dx y)/dx

so you have the eqn:

d(e5x y)/dx=e13x

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