Question

Consider the initial value problem y″+16y=48t,y(0)=3,y′(0)=7. Take the Laplace transform of both sides of the given differential equation to create the corresponding algebraic equation. Denote the Laplace transform of y(t) by Y(s). Do not move any terms from one side of the equation to the other (until you get to part (b) below).

Answer 1

Answer:

$y(t)=3t+sin4t+3cos4t$

Step-by-step explanation:

We are given that initial value problem

$y''+16y=48t$

y(0)=3,y'(0)=7

$L(f^n(t))=s^nL(f(t))-s^{n-1}(f(0)-s^{n-2}f'(0)...-f^{n-1}(0)$

Using the formula then,

$s^2Y(s)-sy(0)-y'(0)+16Y(s)=48L(t)$

$L(t^n)=\frac{n!}{s^{n+1}}$

Substitute the values then, we get

$Y(s)(s^2+16)-3s-7=\frac{48}{s^2}$

$Y(s)(s^2+16)=\frac{48}{s^2}+3s+7$

$Y(s)=\frac{48+3s^3+7s^2}{s^2(s^2+16)}$

$Y(s)=\frac{3}{s^2}+\frac{4}{s^2+16}+\frac{3s}{s^2+16}$

Taking Laplace inverse on both sides then we get

y(t)=$L(\frac{3}{s^2})+L(\frac{4}{s^2+16})+L(\frac{3s}{s^2+16})$

$y(t)=3t+sin4t+3cos4t$

Using formula:$L(\frac{s}{s^2+a^2})=cosat, L(\frac{a}{s^2+a^2})=\frac{1}{a}sinat$

Answer 2

a) The Laplace transform of the entire differential equation is defined by$\mathcal {L}\{f(t)\} = \frac{6}{s^{2}}-\frac{6\cdot s}{s^{2}+16}$.

b) The solution of the given non-homogeneous second order differential equation is $f(t) = 3\cdot t - 6\cdot \cos 4t$.

How to apply Laplace transform in a non-homogeneous differential equation

The Laplace transform is a frequency-based algebraic method used to find a solution associated to linear differential equations in a quick and efficient manner.

In this question we have to find the Laplace transform of the entire differential equation and determine the solution of the differential equation by inverse Laplace transform.

a) We apply the following Laplace transforms to determine the resulting expressions:

$\mathcal {L} \{f(t) + g(t)\} = \mathcal {L}\{f(t)\}+\mathcal{L}\{g(t)\}$   (1)

$\mathcal {L} \{\alpha\cdot f(t)\} = \alpha\cdot \mathcal {L}\{f(t)\}$   (2)

$\mathcal {L} \{y^{(n)} \} = s^{n}\cdot \mathcal {L}\{f(t)\}-s^{n-1}\cdot y(0)-...-y^{(0)}$   (3)

$\mathcal {L} \{t^{n}\} = \frac{n!}{s^{n+1}}$   (4)

$\mathcal {L} \{\sin \omega t\}= \frac{\omega}{s^{2}+\omega^{2}}$   (5)

$\mathcal {L}\{\cos \omega t\} = \frac{\omega}{s^{2}+\omega^{2}}$   (6)

Now we proceed to determine the Laplace transform of the entire differential equation:

$s^{2}\cdot \mathcal {L} \{f(t)\}-s\cdot y(0)-y'(0) + 16\cdot \mathcal {L} \{f(t)\} = 48\cdot \left(\frac{2!}{s^{2}} \right)$

$(s^{2}+16)\cdot \mathcal {L}\{f(t)\} = \frac{96}{s^{2}}+s\cdot y(0) +y'(0)$

$\mathcal {L} \{f(t)\} = \frac{96}{s^{2}\cdot (s^{2}+16)} + \frac{3\cdot s}{s^{2}+16} +\frac{7}{s^{2}+16}$

$\mathcal {L}\{f(t)\} = \frac{6}{s^{2}}-\frac{6\cdot s}{s^{2}+16}$

The Laplace transform of the entire differential equation is defined by$\mathcal {L}\{f(t)\} = \frac{6}{s^{2}}-\frac{6\cdot s}{s^{2}+16}$. $\blacksquare$

b) In this part we shall apply the inverse Laplace transform to find the solution of this differential equation:

$\mathcal {L}\{f(t)\} = \frac{6}{s^{2}}-\frac{6\cdot s}{s^{2}+16}$

$\mathcal {L} \{f(t)\} = 3\cdot \left(\frac{2!}{s^{2}} \right) - 6\cdot \left(\frac{s}{s^{2}+16} \right)$

$f(t) = 3\cdot t - 6\cdot \cos 4t$

The solution of the given non-homogeneous second order differential equation is $f(t) = 3\cdot t - 6\cdot \cos 4t$. $\blacksquare$

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