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I have the following differential equation $\frac{dy}{dx} = 3x+2y+xy$ with $y(0)=-1$ (I don't have the exact solution to this). I want to find the approximate value of $y(0.1)$ using a 6th order Taylor Series. I used Runge-Kutta 4 and it says that $y(0.1) \approx -1.2114$.

What I come up using the Taylor Series method is $\approx -1.23$.

I thought Taylor Series would be more accurate, or maybe I did something wrong in my Taylor Series solution?

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  • $\begingroup$ Where is your Taylor Series solution that you think is wrong? $\endgroup$
    – NoChance
    Feb 28, 2012 at 22:04
  • $\begingroup$ My expansion of the differential equation into a Taylor Series was incorrect. $\endgroup$
    – Zahir J
    Mar 2, 2012 at 14:44

1 Answer 1

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Note that the factorization below suggests a the simpler equation, i.e. translating $(-2,-3)$ to the origin $$ y'=(x+2)(y+3)-6 $$ and the initial condition $(0,-1)$ to $(2,2)$. This, however, leaves $(y+3)'=y'$ unchanged: $$ \eqalign{ y' &= xy - 6 \\ y' - xy &= - 6 \\ (y' - xy) \, e^{-x^2/2} &= - 6\,e^{-x^2/2} \\ (y \, e^{-x^2/2})' &= - 6\,e^{-x^2/2} \\ y \, e^{-x^2/2} &= - 6 \int e^{-x^2/2} dx \\ y \, e^{-x^2/2} &= c-6 \sqrt{\frac\pi2}\, \text{erf}\frac{x}{\sqrt{2}} \qquad \implies \quad c=y_0 \\ y &= e^{x^2/2} \left( y_0 - 3 \sqrt{2\pi}\, \text{erf}\frac{x}{\sqrt{2}}\right) . } $$ Translating back, we get $$ y = e^{(x+2)^2/2} \left( c - 3 \sqrt{2\pi}\, \text{erf}\frac{x+2}{\sqrt{2}}\right)-3 $$ for the same constant $c$ which, however, is no longer equal to $y_0$, but rather to $$ \eqalign{ c &= (y_0+3) e^{-\frac12(x_0+2)^2} + 3\sqrt{2\pi}\,\text{erf}\frac{x_0+2}{\sqrt{2}} \\&=2e^{-2}+3\sqrt{2\pi}\,\text{erf}\sqrt{2} \approx 7.44839864640532 } $$ which would give $$ \eqalign{ y(0.1) &= e^{1/200} \left( c - 3 \sqrt{2\pi}\, \text{erf}\frac{2.1}{\sqrt{2}}\right)-3 \\&\approx -1.21143171766941 \,. } $$ If we had just let $u=x+2$ and $v=y+3$ to begin with, we could have rewritten our differential equation as $v'=uv-6$ which is easier to differentiate by hand anyway: $$ \eqalign{ \\v^{(1)}&=\left(u\right)\cdot&v+\left(-6\right) \\v^{(2)}&=\left(u^{2}+1\right)\cdot&v+\left(-6 \, u\right) \\v^{(3)}&=\left(u^{3}+3\,u\right)\cdot&v+\left(-6\,u^{2}-12\right) \\v^{(4)}&=\left(u^{4}+6\,u^{2}+3\right)\cdot&v+\left(-6\,u^{3}-30\,u\right) \\v^{(5)}&=\left(u^{5}+10\,u^{3}+15\,u\right)\cdot&v +\left(-6\,u^{4}-54\,u^{2}-48\right) \\v^{(6)}&=\left(u^{6}+15\,u^{4}+45\,u^{2}+15\right)\cdot&v +\left(-6\,u^{5}-84\,u^{3}-198\,u\right) \\v^{(7)}&=\left(u^{7}+21\,u^{5}+105\,u^{3}+105\,u\right)\cdot&v +\left(-6\,u^{6}-120\,u^{4}-522\,u^{2}-288\right) \\v^{(8)}&=\left(u^{8}+28\,u^{6}+210\,u^{4}+420\,u^{2}+105\right)\cdot&v +\left(-6\,u^{7}-162\,u^{5}-1110\,u^{3}-1674\,u\right) \\v^{(9)}&=\left(u^{9}+36\,u^{7}+378\,u^{5}+1260\,u^{3}+945\,u\right)\cdot&v +\left(-6\,u^{8}-210\,u^{6}-2070\,u^{4}-5850\,u^{2}-2304\right) } $$ especially if one notices the recursion $$ \left.\matrix{ v^{(n)}=a_nv+b_n\\\\ a_1=u,~b_1=-6}\right\} \quad\implies\quad \left\{\matrix{ a_{n+1}=a_n'+u\,a_n\\\\ b_{n+1}=b_n'-6\,a_n\,,}\right. $$ which I managed to calculate in sage:

var('a,b,u,v,z')
a(u) = u
b(u) = -6
for n in range(1,10):
    # print n, a, b #Note: a & b are functions of u
    # print n, a(u).simplify_full(), b(u).simplify_full()
    print '\\\\v^{(%d)}&=\\left(%s\\right)\cdot v+\\left(%s\\right)' \
    % (n, latex(a(u).simplify_full()), latex(b(u).simplify_full()))
    # print 'v^{(%d)}=%d' % (n, 2*a(2)+b(2))
    z(u) = a
    a(u) = (diff(a,u) + u*z).simplify_full()
    b(u) = (diff(b,u) - 6*z).simplify_full()

Using the initial condition $(u,v)=(2,2)$ corresponding to $(x,y)=(0,-1)$, we get $$ v^{(1)}=-2\\ v^{(2)}=-2\\ v^{(3)}=-8\\ v^{(4)}=-22\\ v^{(5)}=-76\\ v^{(6)}=-262\\ v^{(7)}=-980\\ v^{(8)}=-3794\\ v^{(9)}=-15428 $$ so that, factoring out the $(-1)$, I computed the following terms and partial sums:

t = [1, 2, 2, 8, 22, 76, 262, 980, 3794, 15428]
x = 0.1
y = 0
for n in range(10):
    tn = t[n]*x^n/factorial(n)
    y = y + tn
    print '%-4d %-14g %-16.12g' % (n, tn, y)

0    1              1               
1    0.2            1.2             
2    0.01           1.21            
3    0.00133333     1.21133333333   
4    9.16667e-05    1.211425        
5    6.33333e-06    1.21143133333   
6    3.63889e-07    1.21143169722   
7    1.94444e-08    1.21143171667   
8    9.40972e-10    1.21143171761   
9    4.25154e-11    1.21143171765  

So the analytic and Taylor series solutions agree.

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  • $\begingroup$ I was actually just looking for the Taylor Series expansion (your final step) but you've overdone yourself. Thank you! $\endgroup$
    – Zahir J
    Mar 2, 2012 at 14:46

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