Method of undetermined coefficients

maeveoneill

Junior Member
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Sep 24, 2005
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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?!
 
Let \(\displaystyle \L\\y_{p}=Ax+Bcos(2x)+Csin(2x)\)

Then, \(\displaystyle \L\\y'_{p}=A-2Bsin(2x)+2Ccos(2x)\)

\(\displaystyle \L\\y''_{p}=-4Bcos(2x)-4Csin(2x)\)

Sub them into \(\displaystyle \L\\y''+y'-2y\)

Simplify and use the coefficients of like terms.

Here's what you're working toward:

\(\displaystyle \L\\y=\frac{-cos(2x)}{20}-\frac{3sin(2x)}{20}+\frac{e^{-2x}}{6}+\frac{17e^{x}}{15}-\frac{1}{2}x-\frac{1}{4}\)
 
galactus said:
Let \(\displaystyle \L\\y_{p}=Ax+Bcos(2x)+Csin(2x)\)

Then, \(\displaystyle \L\\y'_{p}=A-2Bsin(2x)+2Ccos(2x)\)

\(\displaystyle \L\\y''_{p}=-4Bcos(2x)-4Csin(2x)\)

Sub them into \(\displaystyle \L\\y''+y'-2y\)

Simplify and use the coefficients of like terms.

Here's what you're working toward:

\(\displaystyle \L\\y=\frac{-cos(2x)}{20}-\frac{3sin(2x)}{20}+\frac{e^{-2x}}{6}+\frac{17e^{x}}{15}-\frac{1}{2}x-\frac{1}{4}\)



Hi, how did you get the first part

Let \(\displaystyle \L\\y_{p}=Ax+Bcos(2x)+Csin(2x)\)
 
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