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 (5 x) dx / (y^{5} + 3)

and put integral signs in to get:

$\int y^{4} / (y^{5} + 3) dy = \int 2 \sin (5 x) dx$

Now just do the integrals (remembering to put the constant in) and substitute $x = π$ $y = 0$ 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 $\int y' / y . dx = \log (y) + C$ . 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 $(y . e^{\int p (x) dx})' = y' e^{\int p (x) dx} + p (x) y e^{\int p (x) dx} = q (x) e^{\int p (x) dx}$ . Integrating both sides of that equation and dividing by $e^{\int p (x) dx}$ you get:

$y (x) = (\int q (x) e^{\int p (x) dx} dx + C) / e^{\int p (x) dx}$ where $C$ is any arbitrary constant.

To solve your question, just use $p (x) = 3 / x$ and $q (x) = e^{- 3 x} / x^{2}$ .

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

So;

$\int y^{4} / (y^{5} + 3) dy = 3 / 5 \log y^{5} + c$ ??

and,

$\int 2 \sin (5 x) dx = 10 \cos (5 x) + c$ then you say substitute $x = π$ and $y = 0$ but $\log 0$ ???
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 (y^{5} + 3) = - 2 \cos (5 x) + C

. So when you substitute

y = 0

x = π

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 $\int 3 / x dx = 3 \log (x)$ (we don't need a constant here). So $e^{\int 3 / x dx} = e^{3 \log (x)} = x^{3}$ . Using the formula I gave, we have that $y (x) = (\int e^{- 3 x} / x^{2} . x^{3} dx + C) / x^{3}$ . So we need to work out $\int x . e^{- 3 x} dx$ . Have you done integration by parts? Integration by parts gives you the formula $\int u . v' . dx = u v - \int u' . v . dx$ where $u$ and $v$ are functions of $x$ . In this case, we set $u = x$ and $v' = e^{- 3 x}$ because we know that if $v' = e^{- 3 x}$ then $v = - 1 / 3 . e^{- 3 x}$ , and $u' = 1$ . So $\int x . e^{- 3 x} dx = - 1 / 3 . x . e^{- 3 x} - \int 1 . - 1 / 3 . e^{- 3 x} dx = (1 / 3) (x e^{- 3 x} + \int e^{- 3 x} dx) = (1 / 3) (x e^{- 3 x} - (1 / 3) e^{- 3 x}) = (1 / 9) (3 x - 1) e^{- 3 x}$ . So we have that $y = ((1 / 9) (3 x - 1) e^{- 3 x} + C) / x^{3}$ .

I hope I got that right.