Posted by **COFFEE** on Thursday, July 12, 2007 at 10:53pm.

Please check my work and correct any errors/point out any errors. Thanks.

Solve the initial-value problem using the method of undetermined coefficients.

y''-4y=e^xcos(x), y(0)=1, y'(0)=2

r^2-4=0, r1=2, r2=-2

yc(x)=c1*e^2x+c2*e^-2x

yp(x)=e^x[Acos(x)+Bsin(x)]

y'p(x)=e^x[-Asin(x)+Bcos(x)]

y"p(x)=e^x[-Acos(x)-Bsin(x)]

e^x[-Acos(x)-Bsin(x)]-4e^x[Acos(x)+Bsin(x)]

A=-1/5,B=1/10

yp(x)=e^x[(-1/5)cos(x)+(1/10)sin(x)]

y(x)=c1*e^2x+c2*e^-2x+e^x[(-1/5)cos(x)+(1/10)sin(x)]

y(0)=c1+c2-1/5=1

c1=6/5-c2

y'(x)=2c1*e^2x-2c2*e^-2x+e^x[(1/5)cos(x)+(1/10)cos(x)]

y'(0)=12/5-2c2-2c2+1/10=2

c2=1/10, c1=11/10

y(x)=(11/10)e^2x+(1/10)e^-2x+e^x[(-1/5)cos(x)+(1/10)sin(x)]

Can someone please answer this question? I have yet to get any help with it.

The line just above A= -1/5, B=1/10

Shouldn't that equal e^xcosx? I don't follow your work.

I'm missing some terms in y'p and y''p (Leibnitz rule, you also have to differentiate the exponential...).

Anyway, to solve problems with less chance of making mistakes, you must cut down on the number of steps you use. In this case you can write:

e^x cos(x) =

1/2 [e^(x(1+i)) + e^(x(1-i))]

Using the superposition principle yo can solve the equations:

y'' - 4 y = 1/2 e^(x(1+i))

and

y'' - 4 y = 1/2 e^(x(1-i))

separately and add them up.

So, all you have to do is solve the equation:

y'' - 4 y = 1/2 e^(a x) ---->

yp = 1/2 e^(ax)/[a^2 - 4]

insert a = 1 + i and a = 1 - i and add the two terms up.

Then you add the homogeneous solution: yh = A e(2x) + B^e^(-2x)

ok thanks!

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