Answer :
To solve the given differential equation using Laplace Transform, let's follow the steps below:
### Part (a): Proving the Form of [tex]\(Y(s)\)[/tex]
Given the differential equation:
[tex]\[ t \frac{d^2 y}{d t^2} + \frac{d y}{d t} + t y(t) = 0 \][/tex]
with initial conditions [tex]\( y(0) = 2 \)[/tex] and [tex]\( \frac{d y}{d t}(0) = 0 \)[/tex].
First, recall some properties of the Laplace Transform:
1. [tex]\( \mathcal{L} \left\{ y'(t) \right\} = s Y(s) - y(0) \)[/tex]
2. [tex]\( \mathcal{L} \left\{ t y'(t) \right\} = -\frac{dY(s)}{ds} \)[/tex]
3. [tex]\( \mathcal{L} \left\{ t \frac{d^2 y}{d t^2} \right\} = -s \frac{dY(s)}{ds} - Y(s) \)[/tex]
Now, taking the Laplace Transform of both sides of the given differential equation:
[tex]\[ t \frac{d^2 y}{d t^2} + \frac{d y}{d t} + t y(t) = 0 \][/tex]
Apply the Laplace Transform:
[tex]\[ \mathcal{L} \left\{ t \frac{d^2 y}{d t^2} \right\} + \mathcal{L} \left\{ \frac{d y}{d t} \right\} + \mathcal{L} \left\{ t y(t) \right\} = 0 \][/tex]
Using the properties of Laplace Transforms mentioned above, we get:
[tex]\[ -s \frac{dY}{ds} - Y(s) + sY(s) - y(0) + \frac{dY(s)}{ds} + t Y(s) = 0 \][/tex]
Simplifying, we obtain:
[tex]\[ -s \frac{dY}{ds} + \frac{dY}{ds} + s Y(s) + Y(s) - y(0) = 0 \][/tex]
Substitute the initial condition [tex]\( y(0) = 2 \)[/tex]:
[tex]\[ -s \frac{dY}{ds} + \frac{dY}{ds} + s Y(s) + Y(s) - 2 = 0 \][/tex]
Combine like terms:
[tex]\[ (1 - s) \frac{dY}{ds} + (s + 1) Y(s) = 2 \][/tex]
Given the form [tex]\( Y(s) = \frac{A}{\sqrt{s^2 + 1}} \)[/tex], let's verify if it fits this differential equation. Assume [tex]\( Y(s) \)[/tex] has the form:
[tex]\[ Y(s) = \frac{A}{\sqrt{s^2 + 1}} \][/tex]
### Part (b): Finding the Value of Constant [tex]\(A\)[/tex]
To determine the constant [tex]\(A\)[/tex], apply the initial conditions to the Laplace-transformed equation.
From initial condition [tex]\( y(0) = 2 \)[/tex]:
[tex]\[ \mathcal{L} \left\{ y(t) \right\}(0) = Y(0) = \frac{A}{\sqrt{0^2 + 1}} \][/tex]
Therefore:
[tex]\[ Y(0) = \frac{A}{1} = A \][/tex]
Since [tex]\( y(0) = 2 \)[/tex]:
[tex]\[ A = 2 \][/tex]
Thus, the value of the constant [tex]\(A\)[/tex] is [tex]\( 2 \)[/tex].
### Conclusion
(a) We have shown that [tex]\( Y(s) \)[/tex] can be written in the form [tex]\( \frac{A}{\sqrt{s^2 + 1}} \)[/tex].
(b) By applying the initial conditions, we determined that [tex]\( A = 2 \)[/tex].
### Part (a): Proving the Form of [tex]\(Y(s)\)[/tex]
Given the differential equation:
[tex]\[ t \frac{d^2 y}{d t^2} + \frac{d y}{d t} + t y(t) = 0 \][/tex]
with initial conditions [tex]\( y(0) = 2 \)[/tex] and [tex]\( \frac{d y}{d t}(0) = 0 \)[/tex].
First, recall some properties of the Laplace Transform:
1. [tex]\( \mathcal{L} \left\{ y'(t) \right\} = s Y(s) - y(0) \)[/tex]
2. [tex]\( \mathcal{L} \left\{ t y'(t) \right\} = -\frac{dY(s)}{ds} \)[/tex]
3. [tex]\( \mathcal{L} \left\{ t \frac{d^2 y}{d t^2} \right\} = -s \frac{dY(s)}{ds} - Y(s) \)[/tex]
Now, taking the Laplace Transform of both sides of the given differential equation:
[tex]\[ t \frac{d^2 y}{d t^2} + \frac{d y}{d t} + t y(t) = 0 \][/tex]
Apply the Laplace Transform:
[tex]\[ \mathcal{L} \left\{ t \frac{d^2 y}{d t^2} \right\} + \mathcal{L} \left\{ \frac{d y}{d t} \right\} + \mathcal{L} \left\{ t y(t) \right\} = 0 \][/tex]
Using the properties of Laplace Transforms mentioned above, we get:
[tex]\[ -s \frac{dY}{ds} - Y(s) + sY(s) - y(0) + \frac{dY(s)}{ds} + t Y(s) = 0 \][/tex]
Simplifying, we obtain:
[tex]\[ -s \frac{dY}{ds} + \frac{dY}{ds} + s Y(s) + Y(s) - y(0) = 0 \][/tex]
Substitute the initial condition [tex]\( y(0) = 2 \)[/tex]:
[tex]\[ -s \frac{dY}{ds} + \frac{dY}{ds} + s Y(s) + Y(s) - 2 = 0 \][/tex]
Combine like terms:
[tex]\[ (1 - s) \frac{dY}{ds} + (s + 1) Y(s) = 2 \][/tex]
Given the form [tex]\( Y(s) = \frac{A}{\sqrt{s^2 + 1}} \)[/tex], let's verify if it fits this differential equation. Assume [tex]\( Y(s) \)[/tex] has the form:
[tex]\[ Y(s) = \frac{A}{\sqrt{s^2 + 1}} \][/tex]
### Part (b): Finding the Value of Constant [tex]\(A\)[/tex]
To determine the constant [tex]\(A\)[/tex], apply the initial conditions to the Laplace-transformed equation.
From initial condition [tex]\( y(0) = 2 \)[/tex]:
[tex]\[ \mathcal{L} \left\{ y(t) \right\}(0) = Y(0) = \frac{A}{\sqrt{0^2 + 1}} \][/tex]
Therefore:
[tex]\[ Y(0) = \frac{A}{1} = A \][/tex]
Since [tex]\( y(0) = 2 \)[/tex]:
[tex]\[ A = 2 \][/tex]
Thus, the value of the constant [tex]\(A\)[/tex] is [tex]\( 2 \)[/tex].
### Conclusion
(a) We have shown that [tex]\( Y(s) \)[/tex] can be written in the form [tex]\( \frac{A}{\sqrt{s^2 + 1}} \)[/tex].
(b) By applying the initial conditions, we determined that [tex]\( A = 2 \)[/tex].
