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

1. ## 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

2. 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

3. 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.