To solve the initial value problem $y'' - 2y' - 3y = 0$, with the initial conditions $y(0) = -4$ and $y'(0) = 12$, using the Laplace transform method:

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Step 1: Apply the Laplace Transform

Taking the Laplace transform of both sides of the differential equation:

$\mathcal{L}\{y''\} - 2\mathcal{L}\{y'\} - 3\mathcal{L}\{y\} = 0$

Using the Laplace transform properties:

• $\mathcal{L}\{y''\} = s^2 Y(s) - sy(0) - y'(0)$,
• $\mathcal{L}\{y'\} = s Y(s) - y(0)$,
• $\mathcal{L}\{y\} = Y(s)$,

Substitute these into the equation:

[ [s^2 Y(s) - s(-4) - 12] - 2[s Y(s) - (-4)] - 3Y(s) = 0 ]

Simplify:

$s^2 Y(s) + 4s + 12 - 2s Y(s) - 8 - 3Y(s) = 0$

$(s^2 - 2s - 3)Y(s) + (4s + 4) = 0$

Rearrange:

$(s^2 - 2s - 3)Y(s) = -(4s + 4)$

$Y(s) = \frac{-(4s + 4)}{s^2 - 2s - 3}$

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Step 2: Factorize the Denominator

Factor $s^2 - 2s - 3$:

$s^2 - 2s - 3 = (s - 3)(s + 1)$

So:

$Y(s) = \frac{-(4s + 4)}{(s - 3)(s + 1)}$

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Step 3: Partial Fraction Decomposition

Decompose:

$\frac{-(4s + 4)}{(s - 3)(s + 1)} = \frac{A}{s - 3} + \frac{B}{s + 1}$

Multiply through by the denominator:

$-(4s + 4) = A(s + 1) + B(s - 3)$

Expand:

$-(4s + 4) = A(s) + A + B(s) - 3B$

Group terms:

$-(4s + 4) = (A + B)s + (A - 3B)$

Equating coefficients:

1. $A + B = -4$,
2. $A - 3B = -4$.

Solve the system:

• From $A + B = -4$, $A = -4 - B$.
• Substitute $A = -4 - B$ into $A - 3B = -4$:

$(-4 - B) - 3B = -4$

$-4 - 4B = -4$

$-4B = 0 \implies B = 0$

Substitute $B = 0$ into $A + B = -4$:

$A + 0 = -4 \implies A = -4$

Thus:

$\frac{-(4s + 4)}{(s - 3)(s + 1)} = \frac{-4}{s - 3} + \frac{0}{s + 1}$

Simplify:

$Y(s) = \frac{-4}{s - 3}$

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Step 4: Inverse Laplace Transform

Take the inverse Laplace transform:

$Y(s) = \frac{-4}{s - 3} \implies y(t) = -4e^{3t}.$

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Final Solution:

$y(t) = -4e^{3t}.$

Would you like me to provide further clarification or verification for any steps?

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Relative Questions:

1. How does the Laplace transform handle initial conditions in differential equations?
2. Why is partial fraction decomposition necessary in Laplace transforms?
3. What are the key properties of the Laplace transform for derivatives?
4. Can this method be used for non-homogeneous differential equations?
5. How would the solution change if the initial conditions were different?

Tip:

Always verify the factorization of the characteristic polynomial carefully; errors here propagate through the entire solution.