Question

Find a formula for Y(t) with Y(0)=1 and draw its graph. What is Y\infty?

Answer 1

Answer:

$(a)\ y(t)\ =\ -2e^{-2t}+3$

$(b)\ y(t)\ =\ 4e^{-2t}-3$

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

$y_c(t)\ =\ C.e^{-2t}$

To find the particular integral, we will write

$y_p(t)\ =\ \dfrac{6}{D+2}$

          $=\ \dfrac{6.e^{0.t}}{D+2}$

           $=\ \dfrac{6}{0+2}$

           = 3

so, the total solution can be given by

$y_(t)\ =\ C.F+P.I$

         $=\ C.e^{-2t}\ +\ 3$

$y_(0)=C.e^{-2.0}\ +\ 3$

but according to question

1 = C +3

=> C = -2

So, the complete solution can be given by

$y_(t)\ =\ -2.e^{-2.t}\ +\ 3$

(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

$y_c(t)\ =\ C.e^{-2t}$

To find the particular integral, we will write

$y_p(t)\ =\ \dfrac{-6}{D+2}$

           $=\ \dfrac{-6.e^{0.t}}{D+2}$

           $=\ \dfrac{-6}{0+2}$

           = -3

so, the total solution can be given by

$y_(t)\ =\ C.F+P.I$

         $=\ C.e^{-2t}\ -\ 3$

$y_(0)\ =C.e^{-2.0}\ -\ 3$

but according to question

1 = C -3

=> C = 4

So, the complete solution can be given by

$y_(t)\ =\ 4.e^{-2.t}\ -3$