Find a solution to y 4y 5y 2e2t using variation of parame

Find a solution to y\" - 4y\' - 5y = 2e^2t using variation of parameters., then answer using method of undetermined coefficients.

Solution

First we solve the homogeneous ode.

y\'\'-4y\'-5y=0

Let, y=exp(kt)

Substituting gives

k^2-4k-5=0

k^2-5k+k-5=0

k=5,-1

So, complementary solution ie solution to homogeneous ode is

yh= a exp(5t)+b exp(-t)

Now we look for a particular solution.

IN variation of parameters method we guess particular solution based on complementar solution as

yp= P(x) exp(5t) +Q(x) exp(-t)

with the condition:

P\' exp(5t)+Q\' exp(-t)=0

yp\'= 5P exp(5t)-Q exp(-t)

yp\'\'=5P\' exp(5t)-Q\' exp(-t)+25 exp(5t)+Q exp(-t)

Subtituting gives

5P\' exp(5t)-Q\' exp(-t)+25 P exp(5t)+Q exp(-t)-20P exp(5t)+4Q exp(-t) -5P exp(5t)- 5Q exp(-t)=2e^{2t}

5P\' exp(5t)- Q\' exp(-t)=2 e^(2t)

P\' exp(5t)+Q\' exp(-t)=0

Hence, Q\'= -P\' exp(6t)

5P\' exp(5t)+P\' exp(5t)= 2 e^(2t)

6P\'= exp(-3t)

Integrating gives

6P= -exp(-3t)/3

P = -exp(-3t)/18

Q\' = -P\' exp(6t)= -exp(3t)/6

Q = -exp(3t)/18

Hence particular solution is

yp=-exp(2t)/18-exp(2t)/18=-exp(2t)/9

So general solution is

y=a exp(5t)+b exp(-t)-exp(2t)/9