Two First Order Differential Equations

By Louise on Monday, August 06, 2001 - 11:29 am:

Could someone please go through the following for me
1. Solve the boundary value problem

((dy/dx)y⁴ )/(sin(5x))=2(y⁵ +3)
where y(pi )=0

2. Find the general solution of the differential equation;

dy/dx+(3/x)y=(e^-3x )/x²

Thanks Louise

By Dan Goodman on Monday, August 06, 2001 - 03:42 pm:

For the first one you can do separation of variables. Multiply both sides by sin(5x)dx/(y⁵+3) and put integral signs in to get:

òy⁴/(y⁵+3) dy=ò2sin(5x)dx

Now just do the integrals (remembering to put the constant in) and substitute x=p y=0 to find the value of the constant. The right hand integral hsould be easy enough, for the left hand integral you need to know that òy ' /y.dx=log(y)+C. Let me know if you still have problems.

In general, if you can get all of the y variables on one side of the equals sign, and all of the x variables on the other side, then this is the method you should use.

For the second one, there is a standard equation for the solution. To solve y ' +p(x) y=q(x), where p and q are any functions of x, you have to notice that (y.e^{òp(x) dx}) ' =y ' e^òp(x)dx+p(x) y e^{òp(x) dx} = q(x)e^{òp(x) dx}. Integrating both sides of that equation and dividing by e^{òp(x) dx} you get:

y(x)=(òq(x)e^{òp(x) dx}dx+C)/e^{òp(x) dx} where C is any arbitrary constant.

To solve your question, just use p(x)=3/x and q(x)=e^-3x/x².

By Louise on Tuesday, August 07, 2001 - 03:06 pm:

So;

òy⁴/(y⁵+3) dy=3/5logy⁵+c??

and,

ò2sin(5x) dx=10cos(5x)+c then you say substitute x=p and y=0 but log0???
I do not understand this bit...

And the second question although i'm familiar with the integrating factor technique i just can't seem to manipulate it could anyone go through it step by step?

Thanks Louise x

By David Loeffler on Tuesday, August 07, 2001 - 05:36 pm:

The first integral is in fact 1/5 log(y⁵ +3) +c, and the second integral is -2/5 cos(5x) + c.
But be careful; they are not the same c's! You in fact only need to put in a constant on one integral (it doesn't matter which) - do you see why?

By Dan Goodman on Tuesday, August 07, 2001 - 08:40 pm:

As David says, you should get something like log(y⁵+3)=-2cos(5x)+C. So when you substitute y=0 x=p you get log(3) rather than log(0). Hope that clears that problem up.

On the second one, it's just an exercise in integration. So, first of all you need to work out ò3/x dx=3log(x) (we don't need a constant here). So e^{ò3/x dx}=e^3log(x)=x³. Using the formula I gave, we have that y(x)=(òe^-3x/x².x³ dx+C)/x³. So we need to work out òx.e^-3x dx. Have you done integration by parts? Integration by parts gives you the formula òu.v ' .dx=u v-òu ' .v.dx where u and v are functions of x. In this case, we set u=x and v ' =e^-3x because we know that if v ' =e^-3x then v=-1/3.e^-3x, and u ' =1. So òx.e^-3x dx=-1/3.x.e^-3x-ò1.-1/3.e^-3x dx=(1/3)(x e^-3x+ òe^-3x dx)=(1/3)(x e^-3x-(1/3)e^-3x)=(1/9)(3x-1)e^-3x. So we have that y=((1/9)(3x-1)e^-3x+C)/x³.

I hope I got that right.