Question

By using Laplace transforms, solve the following differential equations subject to the given initial conditions. $$ y^{\prime \prime}-6 y^{\prime}+9 y=t e^{3 t}, \quad y_{0}=0, \quad y_{0}^{\prime}=5 $$

Short answer

y(t) = \delta(t) + 5te^{3t}

Step-by-step solution

Apply the Laplace Transform

Take the Laplace transform of both sides of the differential equation. Remember that the Laplace transform of a derivative \(y''\) and \(y'\) are: \[ L\{y''(t)\} = s^2 Y(s) - sy(0) - y'(0) \] \[ L\{y'(t)\}= sY(s) - y(0) \] Applying these to \(y'' - 6y' + 9y = te^{3t}\): \[ L\{y'' - 6y' + 9y\} = L\{te^{3t}\} \] Using the initial conditions \(y_0 = 0\) and \(y_0' = 5\), we get: \[ s^2 Y(s) - s \times 0 - 5 - 6(sY(s) - 0) + 9Y(s) = \frac{1}{(s-3)^2} \]

Simplify the Laplace Equation

Combine and simplify terms on the left-hand side of the equation: \[ (s^2 - 6s + 9)Y(s) - 5 = \frac{1}{(s-3)^2} \] Add 5 to both sides to isolate \(Y(s)\): \[ (s-3)^2 Y(s) = \frac{1}{(s-3)^2} + 5 \]

Solve for Y(s)

Divide by \( (s-3)^2 \): \[ Y(s) = 1 + \frac{5}{(s-3)^2} \]

Apply the Inverse Laplace Transform

Now find the inverse Laplace transform of \(Y(s)\): \[ y(t) = L^{-1}\{1\} + L^{-1}\{\frac{5}{(s-3)^2}\} \] \(L^{-1}\{1\} = \delta(t)\) and \(L^{-1}\{\frac{5}{(s-3)^2}\} = 5te^{3t}\): \[ y(t) = \delta(t) + 5te^{3t} \]

Key concepts

Differential Equations

Differential equations involve functions and their derivatives and are fundamental in understanding how variables change over time. In the given exercise, the differential equation is a second-order linear type, which means it involves the second derivative of the function y(t). Such equations are common in physics and engineering.
To solve these, we often use initial conditions, which are values of the function and its derivatives at the start point (t=0). Initial conditions help determine a specific solution out of the infinite possibilities.
Laplace transforms are powerful tools to solve differential equations, especially when dealing with non-homogeneous equations, where the non-zero term (here, \( te^{3t} \) ) appears on the right-hand side.

Initial Conditions

Initial conditions specify the value of the dependent variable (and possibly its derivatives) at the start of the observation. They're crucial because they narrow down the solution to the differential equation.
In our exercise, the initial conditions are \( y(0) = 0 \) and \( y'(0) = 5 \). This means:

• At \( t=0 \), the value of \( y \) is 0.
• At \( t=0 \), the first derivative of \( y \), which can be interpreted as the initial rate of change of \( y \), is 5.

These conditions are plugged into the Laplace-transformed differential equation to solve for \( Y(s) \), making the process efficient and straightforward. They ensure the solution is tailored to the specific scenario described.

Inverse Laplace Transform

The inverse Laplace transform is used to convert the Laplace-transformed function back to the time domain. After applying the Laplace transform to a differential equation (making it algebraic and often simpler), solving it, and isolating \( Y(s) \), we need to find the corresponding time-domain function \( y(t) \).
In our problem, we've derived \( Y(s) = 1 + \frac{5}{(s-3)^2} \). The next step is to find the inverse transforms of each part.

• \( L^{-1}\big\{1\big\} = \delta(t) \), where \( \delta(t) \) is the Dirac delta function representing an impulse at \( t=0 \).
• \( L^{-1}\big\{ \frac{5}{(s-3)^2} \big\} = 5te^{3t} \)

Combining these results gives us \( y(t) = \delta(t) + 5te^{3t} \), which is the unique solution to the original differential equation with the given initial conditions.