Find approximate value of y for $x = 0.2, 0.4$ compare your result with exact values

Mumbai University > First Year Engineering > sem 2 > Applied Maths 2

Marks : 6

Year : 2013

By Taylors method,

$y=y_0+hy'_0+\dfrac {h^2}{2!}y"+\dfrac {h^3}{3!}y"'_0 ----- (1)$

$Let \space , y'=\dfrac {dy}{dx}=2x+y\\ y"=2+y'\\ y"' =y"\\ y^{iv}= y"' \\ At x_0 = 0 \space y_0 = 0 \space h=0.2\\ y_0'=2x_0+y_0=2(0)\leftarrow 0=0\\ y"_0=2+y_0'=2+0=2\\ y_0"'=y_0"=2\\ y_0^{iv}=y_0"'=2 \\ \text {Put h=x in equation (1)} \\ y=0+x(0)+\dfrac {x^2}{2!}\times 2+\dfrac {x^3}{3!}\times 2+\dfrac {x^4}{4!}\times 2..... \\ \therefore y=x^2+\dfrac {x^3}3+\dfrac {x^4}{12}+..... \\ Put, h=x=0.2\\ \therefore y=(0.2)^2, +\dfrac {(0.2)^3}3+\dfrac {(0.2)^2}{12} \\ =0.428\\ At x=0.2 \space y_0=0.0428 \space h= 0.2 \\ y_0'=2x_0+y_0=2(0.2)+0.0428\\ =0.4428\\ y_0"=2+y_0' =2+0.4428=2.4428\\ y"'_0=y_0"=2.4428\\ y_0^{iv}=y_0"' =2.4428\\ from(1)\\ y=0.0428 +(0.2)(0.4428)+\dfrac {(0.2)^2}{2!}(2.4428)+\dfrac {(0.2)^3}{3!}(2.4428)+... \\ =0.18347 \\ \text {Exact solution}\\ \dfrac {dy}{dx}=2x+y \hspace {1.5 cm} \text{which is of the type } \dfrac {dy}{dx}+Py=Q\\ \therefore \dfrac {dy}{dx}-y=2x\\ \text{There P = -1 and Q = 2x which area functions of only x. So given P.E. is linear.} \\ I.F=e^{\int Pdx}\\ =e^{-x}\\ \text{Solution is}\\ Y\times I.F=\int Q\times I.F. \space dx\\ y.e^{-x}=\int 2xe^{-x}dx \\ \text{Integrating by parts we get,} \\ ye^{-x}=2\Bigg[x\Bigg(\dfrac {e^{-x}}{-1}\Bigg)-\int 1\times \dfrac {e^{-x}}{-1}dx+c\Bigg]\\ =2[-xe^{-x}-e^{-x}] + c\\ \therefore ye^{-x}=-2e^{-x}(x+1)+c\\ \therefore y=-2(x+1)+ce^{x}\\ Given, x_0=0,y_0=0\\ \therefore 0=-2(0+1)+c\\ \therefore c=2\\ \therefore y=-2(0.2+1)+2e^{0.2}\\ =0.0428\\ Put \space x=0.4\\ Y_1=-2(0.4+1)+2e^{0.4}\\ =0.1836$