Q17E

Question

Find a particular solution to the differential equation.

y''+4y=8sin(2t)

Step-by-Step Solution

Verified
Answer

The particular solution to the differential equation is yp=-2tcos(2t).

1Step 1: Firstly, write the auxiliary equation of the above differential equation

The given differential equation is:

 y''+4y=8sin(2t)                            (1)


Write the homogeneous differential equation of the equation (1),

 

y''+4y=0

 

The auxiliary equation for the above equation,

 

m2+4=0


2Step 2: Now find the roots of the auxiliary equation

Solve the auxiliary equation,

 

m2+4=0m=±2i

 

The roots of the auxiliary equation are: 

 

m1=2i,   &   m2=-2i

 

The complementary solution of the given equation is;

 yc=c1cos2t+c2sin2t

3Step 3: Use the method of undetermined coefficients to find a particular solution to the differential equation

According to the method of undetermined coefficients, assume, the particular solution of equation (1),


yp=t[Asin(2t)+Bcos(2t)]                    .   .....(2)


Now find the derivative of the above equation,


yp=t[Asin(2t)+Bcos(2t)]yp'=(2At+B)cos(2t)+(A-2Bt)sin(2t)yp''=(4A-4Bt)cos(2t)+(-4B-4At)sin(2t)

 

From the equation (1), Substitute the value of  yp'' and yp in the equation (1),


yp''+4yp=8sin(2t)(4A-4Bt)cos(2t)+(-4B-4At)sin(2t)+4(t[Asin(2t)+Bcos(2t)])=8sin(2t)4Acos(2t)-4Bsin(2t)=8sin(2t)

4Step 4: Final conclusion.

Comparing all coefficients of the above equation;

4A=0  A=0-4B=8  B=-2 


 

Therefore, the particular solution of equation (1),

 yp=t[Asin(2t)+Bcos(2t)]yp=-2tcos(2t)