Use Laplace transforms to solve the following initial value problem.

x double prime minus 9 x equals 5 t; x left parenthesis 0 right parenthesis equals x prime left parenthesis 0 right parenthesis equals 0x′′−9x=5t; x(0)=x′(0)=0

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Part 1

​x(t)equals=enter your response here

​(Type an expression using t as the​ variable.)

To solve the given initial value problem using Laplace transforms, we start by taking the Laplace transform of both sides of the differential equation. The given initial value problem is:

\[ x'' - 9x = 5t \]
\[ x(0) = 0 \]
\[ x'(0) = 0 \]

Taking the Laplace transform of both sides:

\[ \mathcal{L}\{x''\} - 9\mathcal{L}\{x\} = \mathcal{L}\{5t\} \]

Using the properties of Laplace transforms, we have:

\[ \mathcal{L}\{x''\} = s^2 X(s) - sx(0) - x'(0) = s^2 X(s) \]
\[ \mathcal{L}\{x\} = X(s) \]
\[ \mathcal{L}\{5t\} = \frac{5}{s^2} \]

Substituting these into the transformed equation:

\[ s^2 X(s) - 9X(s) = \frac{5}{s^2} \]

Factor out \(X(s)\):

\[ X(s)(s^2 - 9) = \frac{5}{s^2} \]

Solve for \(X(s)\):

\[ X(s) = \frac{5}{s^2(s^2 - 9)} \]

Next, we perform partial fraction decomposition on \( \frac{5}{s^2(s^2 - 9)} \):

\[ \frac{5}{s^2(s^2 - 9)} = \frac{A}{s} + \frac{B}{s^2} + \frac{Cs + D}{s^2 - 9} \]

Let's find the coefficients \(A\), \(B\), \(C\), and \(D\):

\[ \frac{5}{s^2(s^2 - 9)} = \frac{A}{s} + \frac{B}{s^2} + \frac{Cs + D}{(s - 3)(s + 3)} \]

Multiply through by the common denominator \(s^2(s^2 - 9)\):

\[ 5 = A s(s^2 - 9) + B(s^2 - 9) + (Cs + D)s^2 \]

Equate coefficients of like powers of \(s\):

For \(s^3\):
\[ 0 = Cs \quad \Rightarrow \quad C = 0 \]

For \(s^2\):
\[ 0 = As^3 + Ds^2 \quad \Rightarrow \quad D = 0 \]

For \(s^1\):
\[ 0 = -9A + 0 \quad \Rightarrow \quad A = 0 \]

For \(s^0\):
\[ 5 = -9B \quad \Rightarrow \quad B = -\frac{5}{9} \]

So the partial fractions are:

\[ \frac{5}{s^2(s^2 - 9)} = -\frac{5}{9} \left( \frac{1}{s^2 - 9} - \frac{1}{s^2} \right) \]

Taking the inverse Laplace transform term-by-term:

\[ X(s) = -\frac{5}{9} \left( \mathcal{L}^{-1}\left\{ \frac{1}{s^2 - 9} \right\} - \mathcal{L}^{-1}\left\{ \frac{1}{s^2} \right\} \right) \]

\[ X(s) = -\frac{5}{9} \left( \mathcal{L}^{-1}\left\{ \frac{1}{s^2 - 3^2} \right\} - \mathcal{L}^{-1}\left\{ \frac{1}{s^2} \right\} \right) \]

Using the inverse Laplace transforms:

\[ \mathcal{L}^{-1}\left\{ \frac{1}{s^2 - a^2} \right\} = \frac{1}{a} \sinh(at) \]
\[ \mathcal{L}^{-1}\left\{ \frac{1}{s^2} \right\} = t \]

Thus,

\[ x(t) = -\frac{5}{9} \left( \frac{1}{3} \sinh(3t) - t \right) \]

Simplifying, we get:

\[ x(t) = -\frac{5}{27} \sinh(3t) + \frac{5}{9}t \]

So the solution is:

\[ x(t) = -\frac{5}{27} \sinh(3t) + \frac{5}{9}t \]

Do you want more details or have any questions?

Here are some questions you can ask next:
1. Can you show me another example of solving an initial value problem using Laplace transforms?
2. What are the common Laplace transform pairs I should know?
3. How do you perform partial fraction decomposition?
4. Can you explain the properties of the Laplace transform?
5. How do you handle different types of initial conditions with Laplace transforms?

**Tip:** Remember that understanding the properties of the Laplace transform and how to manipulate them is key to solving differential equations efficiently.

Learn how to solve the differential equation x'' - 9x = 5t using Laplace transforms with step-by-step guidance. This problem involves differential equations, Laplace transform properties, and inverse Laplace transforms. Suitable for college-level students and those studying advanced mathematics.

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