Find the general solution of d2ydt2 4 dydt 29 y cos 2tSol

Find the general solution of d^2y/dt^2 + 4 dy/dt + 29 y = cos 2t.

Solution

First we solve the assocaited homogeneous ode which is

y\'\'+4y\'+29y=0

This is a linear homogeneous ode with constant coefficients

So solution is of the form: y=exp(kt)

Substituting y=exp(kt) in associated homogeneous ode gives

k^2+4k+29=0

Solving gives

k=2+5i,2-5i

Hence general soluton to homogeneous ode is

y=exp(2t)( A sin(5t)+B cos(5t)

Now we need find aprticular solution

We make the guess based on inhomogeneous part ie cos(2t)

Guess is: yp=C cos(2t)+D sin(2t)

yp\'=-2C sin(2t)+2D cos(2t)

yp\'\'=-4yp

Substituting gives

-4yp+4*(-2C sin(2t)+2D cos(2t))+29yp=cos(2t)

25yp+4*(-2C sin(2t)+2D cos(2t))=cos(2t)

25(C cos(2t)+D sin(2t))+4*(-2C sin(2t)+2D cos(2t))=cos(2t)

(25C+8D)cos(2t)+(25D-8C)sin(2t)=cos(2t)

We get

25C+8D=1

25D=8C

C=25D/8

25*25D/8+8D=1

(25^2+8^2)D=8

D=8/289

C=25/289

yp=(25cos(2t)+8 sin(2t))/289

Hence general solution is

y=y=exp(2t)( A sin(5t)+B cos(5t)+(25cos(2t)+8 sin(2t))/289