Question

The Laplace transform of \(y(t)\) is \(Y(s), y(0)=2\) and \(y^{\prime}(0)=-1\). Find the Laplace transform of the following expressions: (a) \(y^{\prime}\) (b) \(y^{\prime \prime}\) (c) \(2 y^{\prime \prime}-y^{\prime}+3 y\) (d) \(-y^{\prime \prime}+4 y^{\prime}-6 y\) (e) \(\frac{y^{\prime \prime}}{2}+3 y^{\prime}-y\)

Short answer

Answer: \(L\{y'(t)\} = sY(s) - 2\) (b) Given the Laplace transform of y(t), y(0), and y'(0), what is the Laplace transform of y''(t)? Answer: \(L\{y''(t)\} = s^2Y(s) - 2s + 1\) (c) Given the Laplace transform of y(t), y(0), and y'(0), what is the Laplace transform of \(2y'' - y' + 3y\)? Answer: \(L\{2y'' - y' + 3y\} = 2[s^2Y(s) - 2s + 1] - [sY(s) - 2] + 3Y(s)\) (d) Given the Laplace transform of y(t), y(0), and y'(0), what is the Laplace transform of \(-y'' + 4y' - 6y\)? Answer: \(L\{-y'' + 4y' - 6y\} = -[s^2Y(s) - 2s + 1] + 4[sY(s) - 2] - 6Y(s)\) (e) Given the Laplace transform of y(t), y(0), and y'(0), what is the Laplace transform of \(\frac{y''}{2} + 3y' - y\)? Answer: \(L\{\frac{y''}{2} + 3y' - y\} = \frac{1}{2}[s^2Y(s) - 2s + 1] + 3[sY(s) - 2] - Y(s)\)

Step-by-step solution

Laplace transform of the first derivative

Apply the Laplace transform of the derivative formula: \(L\{y'(t)\} = sY(s) - y(0)\). Substitute the given values for Y(s) and y(0): \(L\{y'(t)\} = sY(s) - 2\) (b) \(y''\)

Laplace transform of the second derivative

Apply the Laplace transform of the second derivative formula: \(L\{y''(t)\} = s^2Y(s) - sy(0) - y'(0)\). Substitute the given values for Y(s), y(0), and y'(0): \(L\{y''(t)\} = s^2Y(s) - 2s + 1\) (c) \(2y'' - y' + 3y\)

Apply linearity property to the expression

Using the linearity property, take Laplace transforms of each individual term and add them: \(L\{2y'' - y' + 3y\} = 2[s^2Y(s) - 2s + 1] - [sY(s) - 2] + 3Y(s)\) (d) \(-y'' + 4y' - 6y\)

Apply linearity property to the expression

Using the linearity property, take Laplace transforms of each individual term and add them: \(L\{-y'' + 4y' - 6y\} = -[s^2Y(s) - 2s + 1] + 4[sY(s) - 2] - 6Y(s)\) (e) \(\frac{y''}{2} + 3y' - y\)

Apply linearity property to the expression

Using the linearity property, take Laplace transforms of each individual term and add them: \(L\{\frac{y''}{2} + 3y' - y\} = \frac{1}{2}[s^2Y(s) - 2s + 1] + 3[sY(s) - 2] - Y(s)\)

Key concepts

Differential Equations

Differential equations are mathematical equations that relate a function with its derivatives. They are extremely useful in modeling real-world problems where rates of change are involved, such as physics, engineering, and biology. There are different types of differential equations, such as ordinary differential equations (ODEs), which involve functions of a single variable, and partial differential equations (PDEs), which involve functions of multiple variables. In the context of the Laplace transform, we typically deal with ODEs.

By using the Laplace transform, we can convert differential equations into algebraic equations. This transformation simplifies the process of solving them. Instead of dealing with derivatives directly, we work with transformed equations, making complex calculus a bit simpler. For example, the problem given involves transforming expressions like the first and second derivatives of a function using known formulas for their Laplace transforms.

Using the Laplace transform helps in systematically handling initial conditions and finding unique solutions, which is a major advantage when working with differential equations. This makes the Laplace transform a powerful tool in solving linear differential equations.

Linearity Property

The linearity property of the Laplace transform is one of its most powerful features. It states that the Laplace transform of a linear combination of functions is the same as the linear combination of the Laplace transforms of the individual functions. Mathematically, if you have functions \( f(t) \) and \( g(t) \), then the Laplace transform of \( af(t) + bg(t) \) is \( aL\{f(t)\} + bL\{g(t)\} \).

This property greatly simplifies the process of solving differential equations. In the example provided, the linearity property was used to calculate the Laplace transform of more complex expressions composed of multiple terms like \(2y'' - y' + 3y\). Each of these terms is transformed individually, and then their results are linearly combined, maintaining the coefficients as they appear in the expression. This approach allows us to handle complex expressions conveniently without manually expanding or manipulating each term individually.

The linearity of the Laplace transform is crucial when tackling problems involving multiple terms, ensuring that the problem-solving process is both systematic and efficient.

Initial Conditions

Initial conditions refer to the specific values that the function and its derivatives take at the beginning of a problem, usually given at time \( t = 0 \). In the context of solving differential equations using the Laplace transform, initial conditions are essential for finding specific solutions to those equations.

For instance, in the given exercise, we have initial conditions \( y(0) = 2 \) and \( y'(0) = -1 \). These conditions are used in the Laplace transform formulas for derivatives. When applying the Laplace transform to a derivative, the initial conditions are incorporated into the transformed expression. For example, the Laplace transform of \( y'(t) \) is \( sY(s) - y(0) \), and the Laplace transform of \( y''(t) \) is \( s^2Y(s) - sy(0) - y'(0) \).

Including initial conditions allows us to obtain particular and not just general solutions, tailoring the solution to fit the specific scenario outlined by the problem. In practical applications, using initial conditions in the transformation process provides solutions that accurately reflect initial states, which is crucial in many fields like control systems and mechanical engineering.