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Find a particular solution to the differential equation.

y''+2y'+4y=111e2tcos3t

Short Answer

Expert verified

The particular solution isyp=e2t(cos3t+6sin3t).

Step by step solution

01

Firstly, write the auxiliary equation of the above differential equation. 

The differential equation is:

y''+2y'+4y=111e2tcos3t â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰â€¦(1)

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

y''+2y'+4y=0

The auxiliary equation for the above equation,

m2+2m+4=0

02

Now find the roots of the auxiliary equation 

Solve the auxiliary equation,

m2+2m+4=0m=-2±22-4(1)(4)2(1)m=-2±-122m=-1±i3

The roots of the auxiliary equation are,

m1=-1+i3, â¶Ä‰â¶Ä‰& â¶Ä‰â¶Ä‰m2=-1-i3

The complementary solution of the given equation is,

yc=e-x(c1cos3x+c2sin3x)

03

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=e2t(A0cos3t+B0sin3t) â¶Ä‰â¶Ä‰â€‰â¶Ä‰â¶Ä‰â€‰â¶Ä‰â¶Ä‰â€‰â¶Ä‰â¶Ä‰â€‰â¶Ä‰â¶Ä‰â€‰â¶Ä‰â¶Ä‰......(2)

Now find the derivative of the above equation,

yp'=2e2t(A0cos3t+B0sin3t)+e2t(-3A0sin3t+3B0cos3t)yp''=4e2t(A0cos3t+B0sin3t)+2e2t(-3A0sin3t+3B0cos3t) â¶Ä‰â¶Ä‰â€‰â¶Ä‰â¶Ä‰â€‰â¶Ä‰â¶Ä‰â€‰â¶Ä‰â¶Ä‰+2e2t(-3A0sin3t+3B0cos3t)+e2t(-9A0cos3t-9B0sin3t)yp''=e2t(-5A0cos3t-5B0sin3t)+e2t(-12A0sin3t+12B0cos3t)

From the equation (1), Substitute the value of yp'', â¶Ä‰yp'and yp in the equation (1),

⇒yp''+2yp'+4yp=111e2tcos3t⇒e2t(-5A0cos3t-5B0sin3t)+e2t(-12A0sin3t+12B0cos3t) â¶Ä‰â¶Ä‰â€‰â¶Ä‰â¶Ä‰+2[2e2t(A0cos3t+B0sin3t)+e2t(-3A0sin3t+3B0cos3t)] â¶Ä‰â¶Ä‰â€‰â¶Ä‰â¶Ä‰+4e2t(A0cos3t+B0sin3t)=111e2tcos3t⇒e2t(3A0cos3t+3B0sin3t)+e2t(-18A0sin3t+18B0cos3t)=111e2tcos3t⇒e2tcos3t(3A0+18B0)+e2tsin3t(3B0-18A0)=111e2tcos3t

04

Final conclusion:

Comparing all coefficients of the above equation;

3A0+18B0=111 â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰â€‰......(3)3B0-18A0=0 â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰â€‰â¶Ä‰â¶Ä‰â€‰â¶Ä‰â¶Ä‰......(4)

Solve the above equations,

(3A0+18B0)=111×618A0+108B0=666-18A0+3B0 â¶Ä‰â¶Ä‰â€‰â¶Ä‰=0111B0=666 â¶Ä‰â¶Ä‰â€‰â¶Ä‰â¶Ä‰â€‰â¶Ä‰â¶Ä‰â€‰â¶Ä‰â¶Ä‰â€‰â¶Ä‰â¶Ä‰â€‰â¶Ä‰â¶Ä‰â€‰Â¯â€‰â¶Ä‰â¶Ä‰â€‰â¶Ä‰â¶Ä‰â€‰â¶Ä‰â¶Ä‰B0=6

Substitute the value ofB0 in the equation (3),

3A0+18(6)=1113A0=3A0=1

Therefore, the particular solution of equation (1),

yp=e2t(A0cos3t+B0sin3t)yp=e2t(cos3t+6sin3t)

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