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

Marks : 6

Year : 2013

$$y=y_0+hy_0^1+\dfrac {h^2}{2!}y_0" +\dfrac {h^3}{3!}y_0"' + ..... $$ $ Let, y^1=\dfrac {dy}{dx}=2y+3e^x\\ y"=2y'+3e^x\\ y"'=2y"+3e^x\\ y^{iv}=2y"' +3e^x\\ At \space x_0=0,y_0=0,h=1\\ y_0^1=2y_0+3e^{x_0}=2(0)+3e^{0}=3\\ y_0'=2y_0^1+3e^{x_0}=2(3)+3=9\\ y_0"'=2y_0" +3e^{x_0} =2(9)+3=21\\ y^{iv}_0=2y'"_0+ 3e^{x_0} =2(21)+3=45\\ from (1)\\ y=0+1\times 3\dfrac {1^2}{2!}\times 9+ \dfrac {1^3}{3!}\times 9 +\dfrac {1^3}{3!}\times 21+ \dfrac {1^3}{4!}\times 45...\\ y=3+\dfrac 92+\dfrac {21}6+\dfrac {45}{24}+.....\\ =12.875\\ At \space x_0=1,y_0=12.875,h=0.1\\ y_0^1=2y_0+3e^{x_0}=2(12.875)+3e^{1}=33.9\\ y_0^{11}=2y_0^1+3e^{x_0}=2(33.9)+3e^{1}=76\\ y'"_0=2y"_0+3e^{x_0}=2(76)+3e^{1}=160\\ y^{iv}_0=2y"'_0+3e^{x_0} =2(160)+ 3e^{1}=328.15\\ \therefore y=12.875+0.1(33.9)+\dfrac {0.1^2}{2!}(76)+\dfrac {0.1^3}{3!}(160)+\dfrac {0.1^4}{4!}(328.15)+.... \\ =16.67$