Question

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

Answer 1

`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'`

Answer 2

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`