Chapter 10: Problem 8
Find the general solution of the equation. $$t y^{\prime}+y=\sqrt{t}, t>0$$
Short Answer
Expert verified
The general solution is \( y = \frac{2}{3} t^{\frac{1}{2}} + \frac{C}{t} \).
Step by step solution
01
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 \( g(t) = \sqrt{t} \).
02
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 \( y^{\prime} + \frac{1}{t} y = \frac{\sqrt{t}}{t} = t^{-rac{1}{2}} \).
03
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, \( \mu(t) = e^{\int \frac{1}{t} \, dt} = e^{\ln t} = t \).
04
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 \( t y^{\prime} + y = t^{\frac{1}{2}} \).
05
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, \( y \cdot t = \int t^{\frac{1}{2}} \, dt = \frac{2}{3} t^{\frac{3}{2}} + C \).
06
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.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
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:
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.
- \( \mu(t) = e^{\int P(t) \, dt} \) which involves integrating the coefficient of \( y \), \( P(t) \).
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:
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 \( t \).
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:
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 \( y' + P(t) y = Q(t) \).