# Solution of Initial Value Problems, Laplace Transform

• cc2203
In summary, the Laplace transform is used to solve a given initial value problem. The integral to solve is the Laplace transform of the derivatives of the function with respect to the variable of interest. When solving for A and C, the expressions for B and D are found as B+D=0 and B\omega^2+4D=0. When solving for B and D, the solutions are found to be B=0 and D=0.

#### cc2203

Use the Laplace transform to solve the given initial value problem.

y"+[w^(2)]y=cos2t, w^(2) does not equal 4; y(0)=1, y'(0)=0

I tried doing the problem, and I got up to Y(s)=[(s^(3)+5s]/[s^(2)+w^(2)],
which hopefully is correct. Now I'm having trouble using the Laplace
transforms to finish solving the problem.

What's the integral u have to solve to revert the Laplace transform...?

Daniel.

Taking the Laplace transform of both sides:

$$\mathcal{L}\{y^{''}+\omega^2 y=Cos[2t]\}$$

yields:

$$s^2f-s+\omega^2f=\frac{s}{s^2+4}$$

with:

$$\mathcal{L}\{y(x)\}=f(s)$$

Solving for f(s) yields:

$$f(s)=\frac{s^3+5s}{(s^2+4)(s^2+\omega^2)}$$

Now, you can reduce this using partial fractions with quadratic denominators:

$$\frac{s^3+5s}{(s^2+4)(s^2+\omega^2)}=\frac{As+B}{(s^2+4)}+\frac{Cs+D}{(s^2+\omega^2)}$$

When solving for A,B,C,and D, you'll come up against a homogeneous system for B and D, and the constraint imposed on $\omega^2$ will force you to make a conclusion about what B and D can be. Try solving it to completion and report the results here. If you don't, I'll wrap it up tomorrow.

Just some wrap-up:

Solving for A and C:

$$A=\frac{1}{\omega^2-4}$$

$$C=\frac{5-\omega^2}{4-\omega^2}$$

The expressions for B and D are:

$$B+D=0$$

$$B\omega^2+4D=0$$

You know, I'm not sure about the part below deciding about B and D: Really, I think B and D HAVE to be zero in this case. Someone can correct me if my rational is not so.

Noting that a homogeneous equation has a non-zero solution iff determinant=0, we get:

$$4-\omega^2=0$$

However, since constraint above restricts such, only the trival solution is allowed. Thus:

$$f(s)=A\frac{s}{s^2+4}+C\frac{s}{s^2+\omega^2}$$

Finally, taking the Laplace transform of both sides:

$$y(t)=ACos[2t]+C(Cos[\omega t])$$

A plot is attached.

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## What is the Laplace transform?

The Laplace transform is a mathematical tool that allows us to solve differential equations by transforming them from the time domain to the frequency domain. It is particularly useful for solving initial value problems, where we know the initial conditions of a system but do not have a closed form solution.

## How does the Laplace transform work?

The Laplace transform takes a function of time and converts it into a function of frequency by integrating the function with a complex exponential. This transformed function can then be manipulated algebraically and inverted back into the time domain to find the solution to the original differential equation.

## What types of problems can the Laplace transform solve?

The Laplace transform is most commonly used to solve initial value problems for linear ordinary differential equations with constant coefficients. It can also be used to solve integral and partial differential equations, as well as systems of differential equations.

## What are the advantages of using the Laplace transform?

The Laplace transform has several advantages over other methods of solving differential equations. It can handle a wide range of problems and is particularly useful for solving problems with discontinuous or piecewise continuous functions. It also allows for the use of algebraic techniques, which can be more efficient and less error-prone than traditional methods.

## Are there any limitations to the Laplace transform?

While the Laplace transform is a powerful tool, it does have some limitations. It can only be used to solve linear differential equations, and the initial conditions must be well-defined. It also requires some knowledge of complex analysis to understand and use effectively.