Solving Differential Equations with Simple Taylor Series Method

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The discussion focuses on using the Taylor series method to solve first and possibly second-order differential equations. Participants highlight its simplicity and effectiveness, particularly for non-linear equations, as it allows for the direct computation of derivatives at initial conditions to construct a Taylor polynomial. An example is provided, demonstrating the method with the equation y" = x^2 - y^2 and initial conditions y(0) = 1, y'(0) = 0, leading to a fifth-order Taylor series solution. There is also interest in comparing the Taylor series results with numerical solutions, suggesting a practical approach using software like Mathematica for derivative calculations. Overall, the Taylor series method is seen as a useful shortcut for solving specific differential equations efficiently.
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Hi,
I was reading this math book once... and it had a method for solving differential equations of 1st (And maybe 2nd? I don't remember) order by using simple Taylor series...
I didn't even have to understand much of what was going on, except that I followed some simple rule and I ended up with an accurate solution.
Anyone know where I can find more information about this method? All of the other ones I'm finding show power series which look fairly tedious to compute.
Thanks.
 
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If, for example, you have a differential equation that says
\frac{d^2y}{dx^2}= f(x,y,\frac{dy}{dx})
with initial conditions y(a)= A, y'(a)= B, you can use the equation directly to find y"(a), differentiate the equation to get a formula for y"'(x) and then evaluate to get y"'(a), differentiate again, etc. so that you can get as many derivatives, evaluated at x= a, as you want and construct the Taylor Polynomial. Especially nice for non-linear equations.

Specific example: y"= x2- y2 with initial conditions y(0)= 1, y'(0)= 0.

Immediately y"(0)= 02-(12)= -1.

y"'= 2x- 2yy' so y"'(0)= 2(0)- 2(1)(0)= 0.

yiv= 2- 2(y')2- 2yy" so
yiv(0)= 2- 2(02)-2(1)(0)= 2.

yv= -4(y')2- 2y'y"- 2yy"' so
yv(0)= -4(02)- 2(0)(-1)- 2(1)(0)= 0

To fifth order,
y(x)= 1- \frac{1}{2}x^2+ \frac{2}{4!}x^4

Of course much past there the derivatives are likely to become unwieldly.
 
Alright, thanks for that. :D
 
Thanks, may work with this a bit. You guys mind? You know, solve the DE numerically, then calculate the Taylor series as Hall suggests, then plot the two and see how they match. Hey Pseudo, why don't you do that, say for the equation Hall used. You can take the easy approach like I would do: Have Mathematica calculate the derivatives and just string them together. :smile:
 
Hahah...
If I had Mathematica, that would probably be what I would do... ;)
I'm still looking into solving these numerically with the improved Euler's method... but I guess I'm a little too impatient to be using numerical methods forever.. :P
Looks like Taylor Series is a little short-cut to solving specific equations, heh.
 

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