Please answer all parts of the question (b) Consider the initial value problem y

Question

Please answer all parts of the question

(b) Consider the initial value problem y"(x) + y'(x) + y(x) = 2x + ln x, y(1) = 2, y'(1) = 3 Find approximate values of y(1.1) and y'(1.1) by using Runge-Kuttar method with step size h = 0.1. Write down the numerical scheme clearly. (c) Consider the initial value problem y''(x) = F(x, y(x), y'(x), y''(x)), y(x_0) = a, y'(x_0) = b, y"(x_0) = c Explain how you would rewrite the above initial value problem so that it can be solved by using a method that solves first-order differential equations.

Explanation / Answer

let y'=z

thus

z' =-z -y + 2x + ln x

yn+1= yn + 1/6 ( k1+ 2k2 + 2k3 + k4 )

zn+1=zn + 1/6 ( g1 + 2g2+2g3+g4 )

k1 = 0.1*3=0.3

g1= 0.1* ( -0.3-2 +2 - ln 1 )=-0.03

k2=0.1*( 3+ 0.3/2)=0.3150

g2=0.1*(-2+0.03/2 - 0.3150 + 2*( 1+ 0.1/2)+ ln(1+0.1/2) ) = -0.0151

k3 = 0.1* ( 3 + 0.3150/2)= 0.3158

g3=0.1*(-2+0.0151/2 - 0.3158+ 2*( 1+ 0.1/2)+ ln(1+0.1/2) ) = -0.0159

k4=0.1*( 3+ 0.3158)=0.3316

g4=0.1*(-2+0.0159 - 0.3316+ 2*( 1+ 0.1)+ ln(1+0.1) ) =  -0.0020

thus

y(1.1) = 2+ 1/6*( 0.3 +2*0.03150 +2*0.3158+0.3316 ) =2.2210

g(1.1)=y'(1.1) =3+(-0.03 - 2*0.0151 - 2* 0.0159-0.002 ) = 2.9060

b)

let y'= g

y''=g'=z

thus

y'=g = f ( x, y ) ------(1) with initial condition y'(x0)=g(x0)=b

and

y''=g'= z = f ( x, y, g ) --------------(2) with initial condition y''(x0)=z(0)=c

and hence

z'= f ( x,y, z, g ) ----------(3)

all equation are linear equation and can be solved by method of first order differential equation

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