Explanations 
Textbooks 
Flashcards 
91影视 AI 
Notes 
Study Plans 
Study Sets 
Find a degree 
Magazine Chapter 1: Problem 5
$$ x y^{\prime}+2 y=3 x, y(1)=5 $$
Short Answer
Expert verified
The solution is \(y = \frac{3}{4} x^2 + \frac{17}{4x^2}\).
Step by step solution
01
Recognize the Form of the ODE
The given differential equation is \(xy' + 2y = 3x\). This is a first-order linear differential equation. We can write it in the standard form \(y' + P(x)y = Q(x)\) by dividing every term by \(x\). This gives us \(y' + \frac{2}{x}y = 3\).
02
Find the Integrating Factor
The integrating factor \(\mu(x)\) is computed using \(\mu(x) = e^{\int P(x) \, dx}\). In this case, \(P(x) = \frac{2}{x}\), so \(\mu(x) = e^{\int \frac{2}{x} \, dx} = e^{2\ln x} = x^2\).
03
Multiply Through by the Integrating Factor
Multiply the entire differential equation by the integrating factor. The equation becomes \(x^2 y' + 2xy = 3x^3\). Observe this transforms the left side into the derivative of the product \((x^2 y)'\).
04
Simplify and Integrate
Recognize the left-side expression as \((x^2 y)'\). The equation now looks like \((x^2 y)' = 3x^3\). Integrate both sides with respect to \(x\): \(\int (x^2 y)' \, dx = \int 3x^3 \, dx\), which simplifies to \(x^2 y = \frac{3}{4} x^4 + C\).
05
Solve for \(y\)
Divide through by \(x^2\) to solve for \(y\): \(y = \frac{3}{4} x^2 + \frac{C}{x^2}\).
06
Use Initial Condition to Find \(C\)
Substitute the initial condition \(y(1) = 5\) into the equation \(y = \frac{3}{4} x^2 + \frac{C}{x^2}\). This gives \(5 = \frac{3}{4} \cdot 1^2 + \frac{C}{1^2}\), resulting in \(5 = \frac{3}{4} + C\). Solving for \(C\), we find \(C = \frac{20}{4} - \frac{3}{4} = \frac{17}{4}\).
07
Write the Final Solution
Substitute \(C = \frac{17}{4}\) back into the equation for \(y\). The final solution to the differential equation is \(y = \frac{3}{4} x^2 + \frac{17}{4x^2}\).
Unlock Step-by-Step Solutions & Ace Your Exams!
-
Full Textbook Solutions
Get detailed explanations and key concepts
-
Unlimited Al creation
Al flashcards, explanations, exams and more...
-
Ads-free access
To over 500 millions flashcards
-
Money-back guarantee
We refund you if you fail your exam.
Over 30 million students worldwide already upgrade their learning with 91影视!
Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Initial Condition
An initial condition in the context of differential equations provides a specific value, or 'starting point,' for the solution of the differential equation. By using this value, you can determine the constant of integration, which appears after solving the differential for a specific solution. In our problem, the initial condition is given as \(y(1) = 5\). This means that when \(x = 1\), the function value, \(y\), is 5.
- This value helps eliminate any ambiguities from the integration process by solving for the constant \(C\).
- With the equation \(y = \frac{3}{4} x^2 + \frac{C}{x^2}\), substituting \(x = 1\) and \(y = 5\) allows us to solve for \(C\).
- Only by determining \(C\) can we get the specific solution that satisfies the given initial condition.
Integrating Factor
An integrating factor is a mathematical function used to transform a linear ordinary differential equation (ODE) into one that is easier to solve. The integrating factor, generally denoted as \(\mu(x)\), effectively turns a non-exact equation into an exact one, allowing us to integrate both sides directly.Here's how it works in our exercise:
- The given ODE is first rewritten in standard form: \(y' + \frac{2}{x}y = 3\).
- The integrating factor is found as \(\mu(x) = e^{\int \frac{2}{x} \, dx} = e^{2\ln x} = x^2\).
- Multiplying through by the integrating factor converts the left side of the ODE into a perfect derivative: \((x^2y)'\).
Standard Form of ODE
A first-order linear differential equation typically has the standard form: \[y' + P(x)y = Q(x)\]Here, \(P(x)\) and \(Q(x)\) are functions of \(x\). The beauty of converting an equation into this form is that it sets the stage for applying the integrating factor.For the given equation, \(xy' + 2y = 3x\), the process involves:
- Dividing every term by \(x\) to simplify and reorganize the terms.
- This operation results in \(y' + \frac{2}{x}y = 3\), making it standardized.
