Problem 8 - Calculus: Early Transcendentals

Question

Find the general solution of the equation. $$t y^{\prime}+y=\sqrt{t}, t>0$$

Step-by-Step Solution

Answer

The general solution is \( y = \frac{2}{3} t^{\frac{1}{2}} + \frac{C}{t} \).

Step 1: Identify the Equation Type

The given differential equation is \( t y^{\prime} + y = \sqrt{t} \). This is a first-order linear differential equation in the standard form \( a(t) y^{\prime} + b(t) y = g(t) \). Here, \( a(t) = t \), \( b(t) = 1 \), and \( g(t) = \sqrt{t} \).

Step 2: Convert to Standard Form

Divide every term in the equation by \( t \) to convert the equation into the standard form \( y^{\prime} + P(t) y = Q(t) \). This gives \( y^{\prime} + \frac{1}{t} y = \frac{\sqrt{t}}{t} = t^{-rac{1}{2}} \).

Step 3: Identify Integrating Factor

The integrating factor \( \mu(t) \) is found using the formula \( \mu(t) = e^{\int P(t) \, dt} \), where \( P(t) = \frac{1}{t} \). Thus, \( \mu(t) = e^{\int \frac{1}{t} \, dt} = e^{\ln t} = t \).

Step 4: Multiply Through by Integrating Factor

Multiply the entire differential equation by the integrating factor \( t \): \( t \cdot y^{\prime} + (t \cdot \frac{1}{t}) y = t \cdot t^{-\frac{1}{2}} \), simplifying to \( t y^{\prime} + y = t^{\frac{1}{2}} \).

Step 5: Integrate Both Sides

Recognizing the left side as the derivative of \( y \cdot t \), we integrate: \( \frac{d}{dt}(y \cdot t) = t^{\frac{1}{2}} \). Thus, \( y \cdot t = \int t^{\frac{1}{2}} \, dt = \frac{2}{3} t^{\frac{3}{2}} + C \).

Step 6: Solve for y

Solve for \( y \) by dividing both sides by \( t \): \( y = \frac{2}{3} t^{\frac{1}{2}} + \frac{C}{t} \). This is the general solution of the differential equation.

Key Concepts

Integrating Factor

An integrating factor is a function that we use to simplify differential equations and make them easier to solve. It is mainly employed for first-order linear differential equations. When we have an equation of the form \[ y' + P(t) y = Q(t), \] the integrating factor \( \mu(t) \) is calculated as:

• \( \mu(t) = e^{\int P(t) \, dt} \) which involves integrating the coefficient of \( y \), \( P(t) \).

By multiplying the entire differential equation by this integrating factor, it transforms the left side into a derivative of a product of functions. This simplifies the process of finding solutions.
For example, if \( P(t) = \frac{1}{t} \), the integrating factor \( \mu(t) \) becomes \( e^{\int \frac{1}{t} \, dt} = e^{\ln t} = t \). This transformation makes solving the equation straightforward.

General Solution

The general solution of a differential equation involves finding a function or set of functions that satisfy the given equation. It typically includes a constant of integration, representing an infinite number of possible solutions.
For first-order linear differential equations, once the integrating factor is determined, it helps in integrating and finding the general solution:

• Multiply the entire differential equation by the integrating factor.
• Recognize the left-hand side as the derivative of a product \( \frac{d}{dt} (\text{function}) \).
• Integrate both sides with respect to the independent variable \( t \).

In our example, the general solution \( y = \frac{2}{3} t^{\frac{1}{2}} + \frac{C}{t} \) was obtained by integrating and solving for \( y \). The constant \( C \) indicates that various solutions exist depending on initial conditions.

Differential Equation

A differential equation is a mathematical equation that relates a function with its derivatives. These equations play a crucial role in scientific disciplines as they describe how quantities change over time or space.
The differential equation \( t y^{\prime} + y = \sqrt{t} \) is a first-order linear differential equation. Here, it is important to:

• Identify the type of differential equation.
• Rewrite it into a recognizable form, such as \( y' + P(t) y = Q(t) \).

Such equations require precise manipulation using mathematical tools like integrating factors to find solutions. The essence of solving differential equations is understanding the transformations needed to simplify and resolve the relationships involving derivatives.