0
16kviews
Using Taylor's series method, obtain the solution of the differential equation $\frac{dy}{dx}=3x+y^2$ with $x_0=0,\,y_0=0$ at $x=0.1$ correct to four decimal places.
1 Answer
1
3.5kviews

$ \frac{dy}{dx} = 3x + y^2, x_0 = 0, y_0 = 1 $

$ \text { Taylor's Series is given by } $

$ y = y_0 + (x-x_0)y_0'+ \frac{(x-x_0)^2} {2!}y_0'' $

$ y' = 3x + y^2 \Longrightarrow = 3x_0 + y_0 ^2 = 3(0) + 1 = 1 $

$ y'' = 3 + 2yy' \Longrightarrow y_0'' = 3 + 2y_0y_0' = 3 + 2(1)(1) = 5 $

$ y''' = 2yy'' + 2y^2 \Longrightarrow y_0''' = wy_0y_o' + 2y_0'^2 = 2(1)(5)+2(1) = 12 $

$ y^{IV} = 2yy''' + 2y'y'' + 4y'y'' $

$\Longrightarrow y_0^{IV} = 2y_0y_0''' + 2y_0'y_0'' + 4 y_0'y_0'' $

$\Longrightarrow = 2(1)(12) + 2(1)(5) + 4(1)(5) = 54 $

$ \therefore y = 1 + x + \frac{x^2}{2!}5 + \frac{x^2}{3!}12 + \frac{x^4}{4!} 54 + ...... $

$\text{ When x = 0.1, y = 1.127225} $

Please log in to add an answer.