# Transcribed Image Text: Let a + 0, b + 0, c # 0, a # 0, B#0 and y # 0 represent any real nonzero constants, such that c > 0, a # B, a y and B # y. Let m >1 represent any positive integer. Transcribed Image Text: 10. Use the method of Laplace transform to solve the initial value problem y” – 2ay’ + (a² + B*)y = 0, y(0) = a, y (0) = aa + bB.

To solve the initial value problem using the method of Laplace transform, we first need to take the Laplace transform of the given differential equation and solve for the Laplace transform of the unknown function y(s).

The Laplace transform of a function f(t) is defined as:

F(s) = L[f(t)] = ∫[0,∞] e^(-st) f(t) dt

Where s is a complex variable and t represents time. In this case, we’ll denote the Laplace transform of y(t) as Y(s).

Taking the Laplace transform of the given differential equation, we have:

s^2 Y(s) – s y(0) – y'(0) – 2as Y(s) + 2a y(0) + a^2 Y(s) + B Y(s) = 0

Since y(0) = a and y'(0) = a + bB, we can substitute these values into the equation:

s^2 Y(s) – s a – a – bB – 2as Y(s) + 2a a + a^2 Y(s) + B Y(s) = 0

Next, we group the terms involving Y(s) and factor out the Y(s) terms:

(Y(s)) (s^2 – 2as + a^2 + B) – (s + a + bB) a = 0

Now, we can solve this equation for Y(s). Dividing both sides by (s^2 – 2as + a^2 + B), we obtain:

Y(s) = (s + a + bB) a / (s^2 – 2as + a^2 + B)

This is the Laplace transform of the unknown function y(t).

To find the inverse Laplace transform and obtain the solution y(t), we need to decompose the fraction into partial fractions. However, before we apply partial fraction decomposition, we need to examine the denominator (s^2 – 2as + a^2 + B) to determine its roots.

The roots of the quadratic equation s^2 – 2as + a^2 + B = 0 can be calculated using the quadratic formula:

s = (2a ± √(4a^2 – 4(a^2 + B))) / 2

s = a ± √(-4B) / 2

Since a, B, and y are all nonzero real constants, we have that 4B < 0, which means the roots of the quadratic equation are complex numbers. Consequently, the denominator cannot be factored into linear factors. In this case, we need to use a different approach called the method of completing the square to simplify the denominator and express it in a more convenient form for partial fraction decomposition.

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