Question

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

Answer 1

Answer:

y =$C_1e^{3t}$ +  $Ate^{3t}+ Bsint + Ccost$

Step-by-step 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 =$C_1e^{3t}$ +  $Ate^{3t}+ Bsint + Ccost$

Answer 2

Answer:

$ye^{-3t} = 8t - \frac{2e^{-3t}}{5}(3sin t + cost)+C'$

Step-by-step explanation:

Given differential equation,

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

$\frac{dy}{dt}-3y = 8e^{3t} + 4sin t$

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

$\frac{dy}{dx}+Py=Q$,

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,

$y\times I.F. = \int I.F.\times Q dt$

$\implies y\times e^{-3t}=\int e^{-3t}\times (8e^{3t} + 4sin t)=\int 8+4e^{-3t} sin tdt$

$y\times e^{-3t} = \int 8 dt + 4\int e^{-3t} sin tdt$

$\implies y\times e^{-3t} = 8t + 4\int e^{-3t} sin tdt------(1)$

Let,

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

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

$I=-\frac{e^{-3t} sint}{3} - \int -\frac{e^{3t} cost}{3} dt+C$

$I=-\frac{e^{-3t} sint}{3} - [\frac{e^{-3t} cost }{9}-\int -\frac{e^{-3t} sint}{9}dt]+C$

$I=-\frac{e^{-3t} sint}{3} - [\frac{e^{-3t} cost }{9}+\frac{1}{9}\int e^{-3t} sint dt]+C$

$I=-\frac{e^{-3t} sint}{3} - [\frac{e^{-3t} cost }{9}+\frac{1}{9} I]+C$ ( from equation (2))

$I+\frac{I}{9}=-\frac{e^{-3t} sint}{3} - \frac{e^{-3t} cost }{9}+C$

$\frac{10}{9}I=\frac{-3e^{-3t} sint-e^{-3t} cost }{9}+C$

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

$I=-\frac{e^{-3t}}{10}(3sin t + cost)+C'$

From equation (1),

$y\times e^{-3t} = 8t - \frac{4e^{-3t}}{10}(3sin t + cost)+C'$

$ye^{-3t} = 8t - \frac{2e^{-3t}}{5}(3sin t + cost)+C'$