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Determine the form of a particular solution for the differential equation. Do not solve

y''+5y'+6y=sint-cos2t

Short Answer

Expert verified

Thus, the answer is:yp=Asint+Bcost+Ccos2t+Dsin2t

Step by step solution

01

Firstly, write the auxiliary equation of the given differential equation

The differential equation is,

y''+5y'+6y=sint-cos2t                    ...1

Write the homogeneous differential equation of equation (1),

y''+5y'+6y=0

The auxiliary equation for the above equation,

m2+5m+6=0

02

Now find the complementary solution of the given equation is,

Solve the auxiliary equation,

m2+5m+6=0m2+3m+2m+6=0mm+3+2m+3=0m+2m+3=0

The roots of the auxiliary equation are,

m1=-2,      m2=-3

The complementary solution of the given equation is,

yc=c1e-2t+c2e-3t

03

Now find the form of a particular solution

The particular solution to equation (1) will be the linear combination of termssint,  cos2t and their independent derivatives.

Therefore, the particular solution of equation (1),

yp=Asint+Bcost+Ccos2t+Dsin2t

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