Question

Use Taylor series method to find a solution of ${\displaystyle \frac{dy}{dx}=xy+1,y\left(0\right)=0}$ X=0.2 taking h=0.1 correct upto 4 decimal places. 

Answer 1

(I)${\displaystyle \frac{dy}{dx}=xy+1,}$         ${\displaystyle x_0=,y_0=0=0, h=0.1}$ 

${\displaystyle f(x,y)=1+xy}$ 

${\displaystyle y\prime =1+xy}$                           ${\displaystyle y\prime 0=1}$  

${\displaystyle y\prime \prime =xy+y}$                          ${\displaystyle y\prime \prime 0=0}$ 

${\displaystyle y\prime \prime \prime =xy+2y\prime }$                     ${\displaystyle y\prime \prime \prime o=2}$ 

Taylor’s series is given by , 

${\displaystyle y\left(0.1\right)=y_0+h.y{\prime }_0+\frac{h^2}{2!}y\prime {\prime }_0+\frac{h^3}{3!}y\prime \prime \prime 0+}$ 

=${\displaystyle 0+0.1\left(1\right)+0+\frac{{\left(0.1\right)}^3}{6}}$    (𝟐) 

${\displaystyle y\left(0.1\right)=0.1003}$ 

(II) ${\displaystyle x_1=0,y_1=0.1003,h=0.1}$ 

${\displaystyle y\prime =1+xy}$                           ${\displaystyle y\prime 0=1.01003}$ 

${\displaystyle y\prime \prime =xy\prime +y}$                        ${\displaystyle y\prime \prime 0=0.201303}$

${\displaystyle y\prime \prime \prime =xy\prime \prime +2y\prime }$                ${\displaystyle y\prime \prime \prime 0=2.0401903}$ 

∴${\displaystyle y\left(0.2\right)=0.1003+1.01003\left(0.1\right)+\frac{{0.1}^2}{2!}\left(0.201303\right)+\frac{{0.1}^3}{6}\left(2.0401903\right)}$ 

∴${\displaystyle y\left(0.2\right)=0202708}$