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

Marks : 6

Year : 2014

$$Let \space y'=\dfrac {dy}{dx}=1-2xy$$ $ y"=-2xy'-2y\\ y"' =-2(x.y"+y')-2y'\\ =-2xy"-4y'\\ y^{iv}=-2(xy"+y")-4y"\\ =-2xy"'-6y"\\ At\space x_0=0,y_0=0\\ y_0^1=1-2x_0y_0=1-2(0)(0)=1\\ y'_0=-2x_0y_0'-2y_0=0-2\times 0=0\\ y"_0=-2x_0y_0"-4y"_0=0-4\times 1=0\\y^{iv}_0=-2x_0y_0"'-6y"_0=0-6\times 0=0\\ Let\space h=x\\ from 1\\ y=0+1x+0+\dfrac {x^3}{3!}(-4)+0+....\\ y(x)=x-(2/3)x^3..... (1)\\ Put \space x=0.2\text{ in equation } (1)\\ y(0.2)=0.2-(2/3)(0.2)^3\\=0.1947\\ Put\space x=0.4\text{ in equation} (2)\\ y(0.4)=0.4-\dfrac 23(0.4)^3=0.3753$