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Find the general solution of yʹ − 3y = 8e3t + 4sin t .

User Solsson
by
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2 Answers

7 votes

Answer:


ye^(-3t) = 8t - (2e^(-3t))/(5)(3sin t + cost)+C'

Explanation:

Given differential equation,


y' - 3y = 8e^(3t) + 4sin t


(dy)/(dt)-3y = 8e^(3t) + 4sin t

Since, the above equation is of the type of linear differential equation


(dy)/(dx)+Py=Q,

In which P = -3, Q =
8e^(3t) + 4sin t,

Thus, the integrating factor,


I.F. = e^(\int (-3) dt)=e^(-3t)

Hence, the solution of the given differential equation would be,


y* I.F. = \int I.F.* Q dt


\implies y* e^(-3t)=\int e^(-3t)* (8e^(3t) + 4sin t)=\int 8+4e^(-3t) sin tdt


y* e^(-3t) = \int 8 dt + 4\int e^(-3t) sin tdt


\implies y* e^(-3t) = 8t + 4\int e^(-3t) sin tdt------(1)

Let,


I=\int e^(-3t) sin tdt-----(2)

Integrating by parts, ( First term = sin t, second term =
e^(-3t) )


I=-(e^(-3t) sint)/(3) - \int -(e^(3t) cost)/(3) dt+C


I=-(e^(-3t) sint)/(3) - [(e^(-3t) cost )/(9)-\int -(e^(-3t) sint)/(9)dt]+C


I=-(e^(-3t) sint)/(3) - [(e^(-3t) cost )/(9)+(1)/(9)\int e^(-3t) sint dt]+C


I=-(e^(-3t) sint)/(3) - [(e^(-3t) cost )/(9)+(1)/(9) I]+C ( from equation (2))


I+(I)/(9)=-(e^(-3t) sint)/(3) - (e^(-3t) cost )/(9)+C


(10)/(9)I=(-3e^(-3t) sint-e^(-3t) cost )/(9)+C


I=(1)/(10)(-3e^(-3t) sint-e^(-3t) cost)+C' ( where, C' = 9C/10 )


I=-(e^(-3t))/(10)(3sin t + cost)+C'

From equation (1),


y* e^(-3t) = 8t - (4e^(-3t))/(10)(3sin t + cost)+C'


ye^(-3t) = 8t - (2e^(-3t))/(5)(3sin t + cost)+C'

User Tybstar
by
7.0k points
2 votes

Answer:

y =
C_1e^(3t) +
Ate^(3t)+ Bsint + Ccost

Explanation:


{y}^(') - 3y = 8e^(3t) + 4 sin t


\frac{\mathrm{d} y}{\mathrm{d} t} - 3y = 8e^(3t)+4sint

writing characteristic equation;

( m - 3 ) y = 0

m = 3


y = C_1e^(3t)

for particular solution


y_p = Ate^(3t)+ Bsint + Ccost

hence general solution becomes

y = C.F + P.I

y =
C_1e^(3t) +
Ate^(3t)+ Bsint + Ccost

User Atanu
by
6.5k points