Question

Determine the solution of the differential equation;

d^2y/dx^2+2dy/dy+2y = x^2+1

given that y=4 when x=0 and dy/dx = 8 when x=0

Answer 1

you can use undetermined coefficient here,
y''+2y'+2y=x^2 +1, solve for Yc=0, and the root m
y''+2y'+2y=0
m^2 +2m+2=0...solving for quadratic roots x1=-1+i and x2=-1-i therefore
Yc=C1e^(-1+i)+C2e^(-1-i) or use Yc=C1(e^-x) cosx+C2(e^-x) sinx
Yc=e^-x(C1 cosx+C2sinx)
for Yp, try Yp=Ax^2 +Bx+C, Y'p=2Ax+B, Y''p=2A...sub them into
y''+2y'+2y=x^2 +1, we get
2A+2(2Ax+B)+2(Ax^2 +Bx+C)=x^2 +1
2A+4Ax+2B+2Ax^2 +2Bx+2C=x^2 +1, taking coefficients we get
A=1/2,,,B=-1,,,,,C=1 therefore
Yp=(1/2)x^2 -x +1 now
Y=Yp+Yc
Y=e^-x(C1 cosx+C2sinx)+(1/2)x^2 -x +1 now taking the initial values

using x=0,y=4
4=e^-0(C1cos0+C2sin0)+0-0+1
4=C1 +1
C1=3
using x=0,y'=8
Y'=e^-x[-C1sinx+C2cosx]+[C1 cosx+C2sinx](e^-x)(-1)+x-1
8=e^0[0+C2]-[3cos0+0]e^0-1
8=C2-3-1
8+4=12=C2
Y=e^(-x) [3cosx+12sinx]+ (x^2)/2 -x+1 ...G.S...ans..

Answer 2

finding the roots m^2 +2m+2=0
m1=-1+i ,,m2=-1-i,,,therefore
Yc=C1e^(-1+i)x +C2e^(-1-i)x
or use euler formula for complex numbers to get
Yc=C1e^-x cosx+C2e^-x sinx
Yc=e^(-x)[C1cosx+C2sinx]

assume Yp=Ax^2 +Bx+C,,Y'p=2Ax+B,,Y''p=2A,,sub into
y''+2y'+2y=x^2 +1
2A+2(2Ax+B)+2(Ax^2 +Bx+C)=x^2 +1 we get A=1/2..B=-1,,C=1
Yp=(1/2)x^2 -x+1
now Y=Yc+Yp
Y=e^(-x)[C1cosx+C2sinx]+(1/2)x^2 -x+1
now using the initial conditions x=0,y=4
4=e^(0)[C1cos0+C2sin0]+(1/2)0^2 -0+1
4=C1+1
C1=3

Answer 3

Y'=e^-x[-C1sinx+C2cosx]-[C1cosx+C2sinx]e^x +x-1....,,use x=0,y'=8
8=e^0[-0+C2cos0]-[C1cos0+0]+0-1
8=C2-C1-1
8=C2-3-1
8+4=12=C2
Y=e(-x)[3cosx+12sinx]+ (x^2)/2 -1 ans...G.S.