URGENT HELP! Initial Value Problem Question-Differential Equations and Euler's Method

Printable View

October 11th 2008, 07:29 PM
spazticbutter

URGENT HELP! Initial Value Problem Question-Differential Equations and Euler's Method

dy/dt = 5 - 3(y^(1/2)) y(0) = 2

The problem wants me to solve the initial value problem (for y I presume) and then use it with Euler's method to approximate the solution at t = 0.5,1,1.5,2,and 2.5. I took the equation as separable and took the integral of 1/(5-3(y^(1/2))) and found
(-10/9)ln l 5 - 3(y^(1/2)) l + (2/9)(5 - 3(y^(1/2))) = t + C.
I don't see how I can solve for y from here.
Did I perhaps do my integration wrong? Is there another way to solve this IVP without separation? Or perhaps I don't need to solve for y?
I know that I can solve for C using the given initial value of y but I don't see how this would help me. I know how to do Euler's method so you don't need to explain how to do that unless I have to do it in some special way to figure this problem out :)
Any help would be appreciated :)
October 12th 2008, 04:33 AM
CaptainBlack

Quote:

Originally Posted by spazticbutter

dy/dt = 5 - 3(y^(1/2)) y(0) = 2

The problem wants me to solve the initial value problem (for y I presume) and then use it with Euler's method to approximate the solution at t = 0.5,1,1.5,2,and 2.5. I took the equation as separable and took the integral of 1/(5-3(y^(1/2))) and found
(-10/9)ln l 5 - 3(y^(1/2)) l + (2/9)(5 - 3(y^(1/2))) = t + C.
I don't see how I can solve for y from here.
Did I perhaps do my integration wrong? Is there another way to solve this IVP without separation? Or perhaps I don't need to solve for y?
I know that I can solve for C using the given initial value of y but I don't see how this would help me. I know how to do Euler's method so you don't need to explain how to do that unless I have to do it in some special way to figure this problem out :)
Any help would be appreciated :)

Are you sure that the question does not want you to solve the IVP using Euler's method for t = 0.5,1,1.5,2,and 2.5?

RonL
October 12th 2008, 04:33 AM
shawsend

1 Attachment(s)

There's a lot going on with this one. Three things:

(1) You can solve it directly by separation of variables as you mentioned. I get:

$3\sqrt{y}+5\ln(5-3\sqrt{y})=-9/2(x+c)+5;\quad c=3\sqrt{2}+5\ln(5-3\sqrt{2}),\; 5-3\sqrt{y}>0$

(2) You can solve it numerically with Euler or other means. Not too hard to set up Euler right? Or if you've done a few Euler's, then you can set it up in Mathematica say for the interval (0,5):

Code:

sol = NDSolve[{Derivative[1][y][t] == 5 - 3*y[t]^(1/2), y[0] == 2}, y, {t, 0, 5}] p1 = Plot[Evaluate[y[t] /. sol], {t, 0, 5}]

Yeah, I know. It's almost a crime to do it that way.

(3) You can extract an explicit expression for y in terms of x in (1) using the Lambert $\textbf{W}$ function. Suppose:

$x=y^ne^y$

Then to solve for y in terms of x, it's:

$y=n\textbf{W}\left[\frac{1}{n}x^{1/n}\right]$

So if I have:

$3\sqrt{y}+5\ln(5-3\sqrt{y})=-9/2(x+c)+5$

Can you show:

$y(x)=\Bigg(5/3+5/3\textbf{W}\bigg[-1/5 e^{-1/5(5-k)}\bigg]\Bigg)^2;\quad k=-9/2(x+c)+5$

However if you're like me you're skeptical about all this. So I plotted the NDSolve solution and the Lambert W solution and superimposed them. The plot is below. Works for me.