maeveoneill
Junior Member
- Joined
- Sep 24, 2005
- Messages
- 93
Solve the differential equatoin or initial-value problem using the method of undetermined coefficients: y'' + y' -2y =x +sin2x, y(0)=1, y'(0)=0
This is waht I have so far.. if anyone can tell me whehter its right and lead me in the right direction in completing it.. PLEASE AND THANKS.
r^2 + r -2=0 --->r1 =-2, r2 =1
because b^2 -4ac for this equation is greater than zero we use the general equatoin yc(x) =c1e^-2x +c2e^x
G(x) =x
yp1(x) =Ax + B
y'p(x) =A
y''p(x) O
sub into inital equation
0 + A -2(Ax+ B) =x
??
G(x) =sin 2x
yp2(x)= Ccos2x + Dsin2x
y'p(x) = -Csin2x + Dcos2x
y''(x) = - Ccos2x - D sin2x
then i am stuck.. any help please?!
This is waht I have so far.. if anyone can tell me whehter its right and lead me in the right direction in completing it.. PLEASE AND THANKS.
r^2 + r -2=0 --->r1 =-2, r2 =1
because b^2 -4ac for this equation is greater than zero we use the general equatoin yc(x) =c1e^-2x +c2e^x
G(x) =x
yp1(x) =Ax + B
y'p(x) =A
y''p(x) O
sub into inital equation
0 + A -2(Ax+ B) =x
??
G(x) =sin 2x
yp2(x)= Ccos2x + Dsin2x
y'p(x) = -Csin2x + Dcos2x
y''(x) = - Ccos2x - D sin2x
then i am stuck.. any help please?!