solve the initial value problem ,by Taylor's method of order $N=3$
$y'(t)=ty(t)+(1-t)e^t,0\le t\le 2,y(0)=1$
with an accuracy of $5 \times10^{-3}$
first we consider the taylor expansion of $e^x$
$ e^x=1+x+\frac{x^2}{2!}+\frac{x^3}{3!}+\frac{x^4}{4!}+....$
for $N=3$ we have
$\large e^x=1+x+\frac{x^2}{2}+\frac{x^3}{6} \\ \large y'=yt+(1-t)(1+t+\frac{t^2}{2}+\frac{t^3}{6}) $
bernoulli equation of the form $y'+py=q\implies y u=\int uq dx,u(x)=e^{\int p }dx$
$ y'+(-t)y=1+t+\frac{t^2}{2}+\frac{t^3}{6}$
$u(t)=e^{\int {-t}}dt=e^{-\frac{t^2}{2}} \\ y(t)e^{-\frac{t^2}{2}}=\int (e^{\frac{t^2}{2}})(1-t)(1+t+\frac{t^2}{2}+\frac{t^3}{6}) dt$
$y(t)= e^{\frac{t^2}{2}}\int(1-t)[ e^{-\frac{t^2}{2}}+t e^{-\frac{t^2}{2}}+\frac{1}{2}t^2e^{-\frac{t^2}{2}}+\frac{1}{6}t^3e^{-\frac{t^2}{2}}]dt\\=e^{\frac{t^2}{2}}\int (e^{-\frac{t^2}{2}}-\frac{1}{2}t^2e^{-\frac{t^2}{2}}-\frac{1}{3}t^3e^{-\frac{t^2}{2}}-\frac{1}{6}t^4e^{-\frac{t^2}{2}})dt$
$=e^{\frac{t^2}{2}}\int (e^{-\frac{t^2}{2}})(1-\frac{1}{2}t^2-\frac{1}{3}t^3-\frac{1}{6}t^4)dt$
$\endgroup$ 83 Answers
$\begingroup$Your question is not clear, so lets use three methods to solve this.
We are given:
$$y'(t)=ty(t)+(1-t)e^t,0\le t\le 2,y(0)=1$$
Method 1 Integrating Factor
We can solve this problem and get a result of:
$$y(t) = e^t$$
Method 2 Taylor Series Method
To solve this using a Taylor series with $N = 3$, we assume a solution of the form:
$$\tag 1 \displaystyle y(t) = \frac{y(0)}{0!} + \frac{y'(0)}{1!}(t-0) + \frac{y''(0)}{2!}(t-0)^2 + \frac{y'''(0)}{3!}(t-0)^3 + \ldots$$
We have:
- $y'(t) = ty(t)+(1-t)e^t$
- $y''(t) = y + ty' -te^t$
- $y'''(t) = y' + y' + ty'' -e^t - te^t$
Now, we evaluate each of these at $t=0$, so we have:
- $y(0) = 1$
- $y'(0) = 1$
- $y''(0) = 1$
- $y'''(0) = 1$
Substituting these values back into $(1)$ yields:
$$\displaystyle y(t) = 1 + t + \frac{t^2}{2!} + \frac{t^3}{3!} + \ldots$$
What do you recognize this series as?
Method 3 Numerical Solution using Taylor Series
Using a numerical approach for Taylor methods of order $n$ (in our case, $N = 3$), we have:
- $w_0 = \alpha$
- $w_{i+1} = w_i + h T^{(n)}(t_i,w_i)$ for each $i = 0, 1, \ldots, N-1$,
- where $T^{(n)}(t_i,w_i) = f(t_i,w_i) + \dfrac{h}{2}f'(t_i,w_i)+ \ldots + \dfrac{h^{n-1}}{n!}f^{(n-1)}(t_i,w_i)$
Can you now proceed with Method 3 (you have all the needed derivatives for this method from Method 2)?
$\endgroup$ 8 $\begingroup$Attempt ${2}$
$y'+(-t)y=(1-t)e^t$
$u=e^{\int -tdt}=e^{-\frac{t^2}{2}}$
$\large ye^{\frac{t^2}{2}}=\int e^{-\frac{t^2}{2}}(1-t)e^tdt\implies ye^{-\frac{t^2}{2}}=\int e^{-\frac{t^2}{2}+t}(1-t)=\int e^{\frac{-t^2+2t}{2}}(1-t)dt$
$y e^{\frac{-t^2+2t}{2}} \approx \int(1+\frac{-t^2+2t}{2}+\frac{(\frac{-t^2+2t}{2})^2}{2!}+\frac{(\frac{-t^2+2t}{2})^3}{3!})(1-t)dt$
$let \\ x=\frac{-t^2+2t}{2},dx=(1-t)dt$
$ye^x\approx\int (1+x+\frac{x^2}{2}+\frac{x^3}{6})dx\approx x+\frac{x^2}{2}+\frac{x^3}{6}+\frac{x^4}{24}+c$
$y(0)=1,t=0 \implies x=0\\1(e^0)=0+c,c=1\\y\approx e^{-x}(x+\frac{x^2}{2}+\frac{x^3}{6}+\frac{x^4}{24}+1) $
$\endgroup$ $\begingroup$In fact you don't need to solve this ODE BY Taylor's method, since this ODE has close-form solution:
$y'=ty+(1−t)e^t$
$y'-ty=(1−t)e^t$
I.F. $=e^{\int-t~dt}=e^{-\frac{t^2}{2}}$
$\therefore\left(e^{-\frac{t^2}{2}}y\right)'=(1−t)e^{t-\frac{t^2}{2}}$
$e^{-\frac{t^2}{2}}y=\int(1−t)e^{t-\frac{t^2}{2}}~dt$
$e^{-\frac{t^2}{2}}y=\int(1−t)e^{-\frac{t^2-2t}{2}}~dt$
$e^{-\frac{t^2}{2}}y=\int(1−t)e^{-\frac{t^2-2t+1-1}{2}}~dt$
$e^{-\frac{t^2}{2}}y=\int(1−t)e^{-\frac{(t-1)^2-1}{2}}~dt$
$e^{-\frac{t^2}{2}}y=e^{-\frac{(t-1)^2-1}{2}}+C$
$e^{-\frac{t^2}{2}}y=C+e^{t-\frac{t^2}{2}}$
$y=Ce^\frac{t^2}{2}+e^t$
$y(0)=1$ :
$C+1=1$
$C=0$
$\therefore y=e^t$
$\endgroup$
