Solve the following IVPsSolutionb y 3y 28y 0 The characte

Solve the following IVPs.

Solution

(b) y\" - 3y\' - 28y = 0

The characteristic equation is:

r² - 3r - 28 = 0

Solving it, we get

r = 7 or r = -4

So, the general solution is given by

y(t) = c₁e^7t + c₂ e^-4t

Differentiating both the sides of the above equation, we get

y\'(t) = 7 c₁e^7t - 4 c₂ e^-4t

Putting in the initial conditions,

y(0) = 4 = c₁+ c₂ .....(1)

y\'(0) = 1 = 7 c₁ - 4 c₂ ..(2)

Solving (1) and (2), we get

c₁ = 17 / 11

c₂ = 27 / 11

Hence, the solution is,

y(t) = (17 / 11) e^7t + (27 / 11) e^-4t

(c) y\" + 7y\' + 10y = 0

The characteristic equation is:

r² + 7r + 10 = 0

Solving it, we get

r = -5 or r = -2

So, the general solution is given by

y(t) = c₁e^-5t + c₂ e^-2t

Differentiating both the sides of the above equation, we get

y\'(t) = (-5) c₁e^-5t + (-2) c₂ e^-2t

Putting in the initial conditions,

y(0) = -5 = c₁+ c₂ .....(1)

y\'(0) = 4 = -5 c₁ - 2 c₂ ..(2)

Solving (1) and (2), we get

c₁ = 2

c₂ = - 7

Hence, the solution is,

y(t) = 2 e^-5t - 7 e^-2t

(d) y\" - 9y\' + 18y = 0

The characteristic equation is:

r² - 9r + 18 = 0

Solving it, we get

r = 3 or r = 6

So, the general solution is given by

y(t) = c₁e^3t + c₂ e^6t

Differentiating both the sides of the above equation, we get

y\'(t) = (3) c₁e^3t + (6) c₂ e^6t

Putting in the initial conditions,

y(0) = - 1 = c₁+ c₂ .....(1)

y\'(0) = 8 = 3 c₁ + 6 c₂ ..(2)

Solving (1) and (2), we get

c₁ = -20 / 9

c₂ = 11 / 9

Hence, the solution is,

y(t) = (-20/9) e^3t + (11/9) e^6t