3
$\begingroup$

Given $\frac{dy}{dx}=x^2+y^2$ and initial condition $\varphi (0)=1$, use the method of reduction to an integral equation and successive approximation to find the first 6 terms in the Taylor expansion solution $y=\varphi (x)$.

We have that if $f:D \rightarrow \mathbb{R}$ is continuous, $\varphi$ is defined and continuous on $I=\{x|x_0-h<x<x_0+h\}$ to $\mathbb{R}$, and $(x_0,y_0) \in D$ with $\varphi (x_0)=y_0$, then $\varphi$ is a solution of $\frac{d\varphi}{dx}=f[x,\varphi (x)]$ on $I$ only if $\varphi (x)=y_0+\int_{x_0}^{x} f[t,\varphi (t)] dt$ for $x \in I$. This is what I mean by reduction to an integral equation.

EDIT: In case it helps anyone refine their answer, the answer should be $1+x+x^2+\frac{4}{3}x^3+\frac{7}{6}x^4+\frac{6}{5}x^5$. This comes out of the back of my book. Of course, my problem is that I cannot get to this answer.

$\endgroup$
2
$\begingroup$

$$y = 1 + \int_0^x y^2 + t^2 \, dt$$ we will define $$y_{n+1} = 1 + \int_0^x (t^2 + y_n^2)\, dt = 1 + \frac 13 x^3 + \int_0^x y_n^2 \, dt,\, y_0 = 1.$$ so $$y_1 = 1+ \frac 13 x^3 + \int_0^x \,dt = 1 + x + \frac 13 x^3 \tag 1\\ y_2 = 1 + \frac 13 x^3 + \int_0^x \left(1 + t + \frac 13 t^3\right)^2\, dt = 1 + \frac 13 x^3 + \int_0^x \left(1 + 2t + t^2 + \frac 23 t^3 + \frac 23 t^4 + \frac 19 t^6\right)$$ $$y_2 = 1 + x+ x^2 + \frac23 x^3+\frac 1{6}x^4+\frac 2{15}x^5+\frac 1{63}x^7\tag 2\\ y_2^2 = 1 + 2x + 3x^2 + \frac{10} 3 x^3+\frac 83 x^4 + \cdots\\ y_3 = 1 + \frac 13 x^3 + \int_0^x \left(1 + 2t + 3t^2 + \frac{10} 3 t^3+\frac 83 t^4 + \cdots \right)dt $$ $$y_3=1 + x+x^2 + \frac 43 x^3+\frac 5{17}x^4 + \frac 8{15}x^5+\cdots\tag 3\\ y_3^2 = 1 + 2x + 3x^2 + \frac{14}3x^3+\cdots$$ $$y_4 = 1 + \frac 13 x^3 + \int_0^x \left(1 + 2t + 3t^2 + \frac{14} 3 t^3+ \cdots \right)dt$$ $$ \\y_4 = 1 + x + x^2 + \frac43x^3 + \frac 76 x^4 + \cdots \tag 4 $$

$\endgroup$
  • $\begingroup$ I edited my question to include the expected answer from the back of my book. It seems like you may have gone wrong somewhere. $\endgroup$ – ztforster Mar 25 '15 at 6:30
  • $\begingroup$ Does not seem to be converging to the real answer. I doubt we can use such technique without guaranteeing the convergence of the sequence $y_{n}$ to the actual solution. $\endgroup$ – Paramanand Singh Mar 25 '15 at 7:46
  • $\begingroup$ @ParamanandSingh, please look up picard's theorem. $\endgroup$ – abel Mar 25 '15 at 14:35
  • $\begingroup$ @abel: you should add some details showing that conditions of Picard theorem are satisfied. And I also see that you have corrected the mistake in previous calculation. +1 for you. $\endgroup$ – Paramanand Singh Mar 25 '15 at 16:03
1
$\begingroup$

This is more easily done by repeated differentiation rather than inverting the equation to an integral form and using successive approximation.

The Taylor's series about point $0$ is given by $$\phi(x) = \phi(0) + \phi'(0)x + \frac{\phi''(0)}{2!}x^{2} + \cdots$$ To get six terms we need to calculate derivatives till $\phi^{(6)}(x)$.

Clearly we have $\phi(0) = 1$ and $\dfrac{dy}{dx} = x^{2} + y^{2}$ so that putting $x = 0, y = 1$ we get $\phi'(0) = 0^{2} + 1^{2} = 1$. Again further differentiation gives $$y'' = 2x + 2yy'$$ and so $\phi''(0) = 2\cdot 0 + 2\cdot 1\cdot 1 = 2$. Next we have $$y''' = 2 + 2y'^{2} + 2yy''$$ so that $\phi'''(0) = 2 + 2 + 4 = 8$. Similarly $$y^{(4)} = 6y'y'' + 2yy'''$$ and $\phi^{(4)}(0) = 28$. Again $$y^{(5)} = 6y''^{2} + 8y'y''' + 2yy^{(4)}$$ and $\phi^{(5)}(0) = 144$. Finally $$y^{(6)} = 20y''y''' + 10y'y^{(4)} + 2yy^{(5)}$$ and $\phi^{(6)}(0) = 888$. We then have $$y = \phi(x) = 1 + x + x^{2} + \frac{4}{3}x^{3} + \frac{7}{6}x^{4} + \frac{6}{5}x^{5} + \frac{37}{30}x^{6} + \cdots$$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.