Consider the differential equation y8y12y4ex where primes in

Consider the differential equation

y+8y+12y=4e^(x)

(where primes indicate derivatives with respect to x).

Find a particular solution to the differential equation having the form

yp=u1y1+u2y2

using the method of Variation of Parameters. In this, (Note that this asks for u1 and u2, not u1and u2!)

y1= e^(-6x)

y2= e^(-2x)

u1= -e^(7x)

u2= e^(3x)

Then use your solution to find a solution to the differential equation that satisfies the initial conditions y(0)=0, y(0)=0

y= [BLANK]

Solution

Associated homogeneous ode needs to be solved first which is

y\'\'+8y\'+12y=0

Let,y=exp(kx)

So, k^2+8k+12=0

k=-2,-6

So,y=a exp(-2x)+b exp(-6x)

Particular solution is

yp=u1 exp(-2x)+u2 exp(-6x)

Constraint is

u1\' exp(-2x)+u2\'exp(-6x)=0\\\\

u2\'=-u1\' exp(4x)

yp\'=-2u1 exp(-2x)-6 u2 exp(-6x)

yp\'\'=-2u1\' exp(-2x)-6 u2\' exp(-6x)+4u1 exp(-2x)+36 u2 exp(-6x)

yp\'\'+8yp\'+12yp

=-2u1\' exp(-2x)-6 u2\' exp(-6x)+4u1 exp(-2x)+36 u2 exp(-6x)-16u1 exp(-2x)-48 u2 exp(-6x)+

12u1 exp(-2x)+12u2 exp(-6x)=4e^x

exp(-2x) and exp(-6x) are solutions to homogeneous ode so we get

-2u1\' exp(-2x)-6 u2\' exp(-6x)=4e^x

-u1\'exp(-2x)-3u2\' exp(-6x)=4e^x

u2\'=-u1\' exp(4x)

-u1\'exp(-2x)+3u1\'exp(-2x)=4e^x

2u1\' exp(-2x)=4e^x

u1\'=e^{3x}

Integrating gives

u1=exp(3x)/3

u2\'=-u1\' exp(4x)=-exp(7x)

Integarting gives

u2=-exp(7x)/7

So, yp=exp(x)/3- exp(x)/7=4exp(x)/21

General solution is

y=a exp(-2x)+b exp(-6x)+4exp(x)/21

y(0)=0=a+b+4/21

y\'(0)=-2a-6b+4/21=0

Solving gives

a=-1/3,b=1/7