Question

The function \(y_{P}(t)\) is a particular solution of the given differential equation. Determine the function \(g(t)\). $$y^{\prime \prime}-2 y^{\prime}=g(t), \quad y_{P}(t)=3 t+\sqrt{t}, \quad t>0$$

Short answer

Question: Given the particular solution \(y_P(t) = 3t + \sqrt{t}\) to the second-order linear differential equation \(y^{\prime \prime} - 2y^{\prime} = g(t)\), find the function \(g(t)\). Answer: \(g(t) = \frac{-1-24t + 4\sqrt{t}}{4\sqrt{t}}\)

Step-by-step solution

Find the first derivative of the particular solution

Differentiate the given function \(y_P(t) = 3t + \sqrt{t}\) with respect to \(t\) to get the first derivative \(y_P^{\prime}(t)\). To do this, we apply the power rule to the square root, rewriting \(\sqrt{t}\) as \(t^{1/2}\): $$ y_P^{\prime}(t) = \frac{d}{dt}(3t + t^{1/2}) = 3 + \frac{1}{2}t^{-1/2} $$

Find the second derivative of the particular solution

Differentiate \(y_P^{\prime}(t)\) with respect to \(t\) again to get the second derivative \(y_P^{\prime \prime}(t)\). Once again, apply the power rule: $$ y_P^{\prime \prime}(t) = \frac{d}{dt}\left(3 + \frac{1}{2}t^{-1/2}\right) = -\frac{1}{4}t^{-3/2} $$

Plug the derivatives into the given differential equation

Now, we can substitute the first and second derivatives into the given differential equation \(y^{\prime \prime} - 2y^{\prime} = g(t)\): $$ (-\frac{1}{4}t^{-3/2}) - 2(3 + \frac{1}{2}t^{-1/2}) = g(t) $$

Simplify and find the function \(g(t)\)

Simplify the expression for \(g(t)\) and rewrite it in the form of a single fraction: $$ g(t) = -\frac{1}{4}t^{-3/2} - 6 - t^{-1/2} = \frac{-1-24t + 4\sqrt{t}}{4\sqrt{t}} $$ The function \(g(t)\) is found to be: $$ g(t) = \frac{-1-24t + 4\sqrt{t}}{4\sqrt{t}} $$

Key concepts

Particular Solution

In the world of differential equations, a particular solution is a specific solution that satisfies not only the differential equation but also given conditions or constraints. Unlike the general solution that includes constants representing a family of solutions, a particular solution provides us with a unique answer.
In our case, the function \( y_{P}(t) = 3t + \sqrt{t} \) is given as a particular solution. This function precisely satisfies the differential equation \( y^{\prime\prime} - 2y^{\prime} = g(t) \) for the problem at hand. To find \( g(t) \), we use this particular solution along with its derivatives.
Understanding particular solutions is crucial because it helps us solve real-world problems where specific conditions must be met.

First Derivative

The first derivative of a function gives us the rate at which the function's value is changing at any point, essentially representing the slope of the tangent line to the curve of the function. When finding the first derivative of the particular solution \( y_{P}(t) = 3t + \sqrt{t} \), we apply the basic rules of differentiation, primarily the power rule.
We rewrite \( \sqrt{t} \) as \( t^{1/2} \) to simplify the application of the rule. Applying the power rule, we differentiate to get:
\[ y_{P}^{\prime}(t) = \frac{d}{dt}(3t + t^{1/2}) = 3 + \frac{1}{2}t^{-1/2}. \]
This derivative helps us understand how \( y_{P}(t) \) changes instantaneously, and it is also used to determine the function \( g(t) \) in the differential equation.

Second Derivative

The second derivative of a function provides information on how the rate of change itself is changing. It is the derivative of the first derivative and indicates the concavity or the acceleration of the initial function.
For function \( y_{P}(t) = 3t + \sqrt{t} \), once we have the first derivative \( y_{P}^{\prime}(t) = 3 + \frac{1}{2}t^{-1/2} \), we further differentiate to determine:\
\[ y_{P}^{\prime\prime}(t) = \frac{d}{dt}\left(3 + \frac{1}{2}t^{-1/2}\right) = -\frac{1}{4}t^{-3/2}. \]
This second derivative is a vital component in solving the differential equation since it helps us accurately substitute back to find \( g(t) \). Using the second derivative alongside the first derivative is a common practice in solving higher-order differential equations.

Power Rule

The power rule is a fundamental tool in calculus used for differentiating functions of the form \( x^n \). It states that if \( f(x) = x^n \), then the derivative \( f'(x) = nx^{n-1} \). This rule simplifies the process of finding derivatives and is particularly useful when dealing with polynomial functions.
In the given problem, the power rule is repeatedly used to differentiate the term \( \sqrt{t} \), which we rewrite as \( t^{1/2} \). By applying the power rule, we find:
\[ \frac{d}{dt}(t^{1/2}) = \frac{1}{2}t^{-1/2}, \]
and for the second differentiation,
\[ \frac{d}{dt}(\frac{1}{2}t^{-1/2}) = -\frac{1}{4}t^{-3/2}. \]
Recognizing where and how to apply the power rule can drastically simplify your calculations and deepen your understanding of derivatives across different contexts.