# Math/Calculus

posted by on .

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

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))

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!