Answer:
[tex](a)\ y(t)\ =\ -2e^{-2t}+3[/tex]
[tex](b)\ y(t)\ =\ 4e^{-2t}-3[/tex]
Step-by-step explanation:
(a) Given differential equation is
Y'+2Y=6
=>(D+2)y = 6
To find the complementary function, we will write
D+2=0
=> D = -2
So, the complementary function can be given by
[tex]y_c(t)\ =\ C.e^{-2t}[/tex]
To find the particular integral, we will write
[tex]y_p(t)\ =\ \dfrac{6}{D+2}[/tex]
[tex]=\ \dfrac{6.e^{0.t}}{D+2}[/tex]
[tex]=\ \dfrac{6}{0+2}[/tex]
= 3
so, the total solution can be given by
[tex]y_(t)\ =\ C.F+P.I[/tex]
[tex]=\ C.e^{-2t}\ +\ 3[/tex]
[tex]y_(0)=C.e^{-2.0}\ +\ 3[/tex]
but according to question
1 = C +3
=> C = -2
So, the complete solution can be given by
[tex]y_(t)\ =\ -2.e^{-2.t}\ +\ 3[/tex]
(b) Given differential equation is
Y'+2Y=-6
=>(D+2)y = -6
To find the complementary function, we will write
D+2=0
=> D = -2
So, the complementary function can be given by
[tex]y_c(t)\ =\ C.e^{-2t}[/tex]
To find the particular integral, we will write
[tex]y_p(t)\ =\ \dfrac{-6}{D+2}[/tex]
[tex]=\ \dfrac{-6.e^{0.t}}{D+2}[/tex]
[tex]=\ \dfrac{-6}{0+2}[/tex]
= -3
so, the total solution can be given by
[tex]y_(t)\ =\ C.F+P.I[/tex]
[tex]=\ C.e^{-2t}\ -\ 3[/tex]
[tex]y_(0)\ =C.e^{-2.0}\ -\ 3[/tex]
but according to question
1 = C -3
=> C = 4
So, the complete solution can be given by
[tex]y_(t)\ =\ 4.e^{-2.t}\ -3[/tex]
