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Magazine Chapter 1: Problem 6
$$x^{2} y^{\prime}+3 x y=1 / x, \quad y(1)=-1$$
Short Answer
Expert verified
The solution to the differential equation is \( y = \frac{1}{x^2} - \frac{2}{x^3} \).
Step by step solution
01
Rewrite the Equation
The given differential equation is \(x^2 y' + 3xy = \frac{1}{x}\). We can start by organizing it in the standard form for a first-order linear differential equation. Divide every term by \(x^2\) to start rewriting it as\[ y' + \frac{3}{x} y = \frac{1}{x^3}. \]
02
Identify Components for Integration
The equation \( y' + \frac{3}{x} y = \frac{1}{x^3} \) can now be solved using an integrating factor. Identify \(P(x) = \frac{3}{x}\), and \(Q(x) = \frac{1}{x^3}\).
03
Compute the Integrating Factor
Calculate the integrating factor \(\mu(x)\) as follows:\[ \mu(x) = e^{\int P(x) \, dx} = e^{\int \frac{3}{x} \, dx} = e^{3\ln |x|} = |x|^3. \] Since we are dealing with \(x > 0\), we can use \(x^3\), simplifying the integrating factor.
04
Multiply through by the Integrating Factor
Multiply each term of the differential equation \( y' + \frac{3}{x} y = \frac{1}{x^3} \) by the integrating factor \(x^3\), giving us:\[ x^3 y' + 3x^2 y = 1. \]
05
Recognize and Integrate the Left-hand Side
Notice that the left-hand side is now the derivative of \((x^3 y)\). Thus, the equation becomes:\[ \frac{d}{dx}(x^3 y) = 1. \]Integrate both sides with respect to \(x\):\[ x^3 y = x + C. \]
06
Solve for \(y\)
Solve for \(y\) by dividing both sides by \(x^3\):\[ y = \frac{x + C}{x^3} = \frac{1}{x^2} + \frac{C}{x^3}. \]
07
Use Initial Condition to Find \(C\)
We apply the initial condition \(y(1) = -1\). Substitute \(x = 1\) and \(y = -1\):\[ -1 = \frac{1}{1^2} + \frac{C}{1^3}. \]\[ -1 = 1 + C. \]Solve for \(C\):\[ C = -2. \]
08
Write the Particular Solution
Substitute \(C = -2\) back into the general solution:\[ y = \frac{1}{x^2} - \frac{2}{x^3}. \] This represents the particular solution satisfying the given initial condition.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Integrating Factor
The integrating factor is like a magical multiplier that helps us solve linear differential equations in regular form. It transforms these equations into an easier-to-solve form.
Starting with the reformatted equation from our problem, we have:
Starting with the reformatted equation from our problem, we have:
- Linear form: \(y' + P(x) y = Q(x)\)
- Here, \(P(x) = \frac{3}{x}\) and \(Q(x) = \frac{1}{x^3}\)
- \(\mu(x) = e^{\int P(x) \, dx}\)
- \(\mu(x) = x^3\)
- The equation now looks like a derivative! \(\frac{d}{dx}(x^3y)=1\)
Initial Conditions
Initial conditions are the starting points or constants that allow us to find a specific solution to a differential equation from a family of solutions. They are crucial in transforming general solutions into particular solutions.
In our example, we have the initial condition \(y(1)=-1\). This means:
In our example, we have the initial condition \(y(1)=-1\). This means:
- At \(x=1\), the value of function \(y(x)\) is \(-1\).
- \(x^3 y = x + C\)
- \(-1 = 1 + C\)
- Thus, \(C = -2\)
- \(y = \frac{1}{x^2} - \frac{2}{x^3}\)
Linear Differential Equations
Linear differential equations are a fundamental type of differential equation, representing various natural and human-made phenomena. They have a standard form and specific method for finding solutions, making them a key study area in mathematics.
These equations are considered linear when every term in the equation is of first-degree power, meaning:
These equations are considered linear when every term in the equation is of first-degree power, meaning:
- The dependent variable \(y\) and its derivatives \(y'\), \(y''\), etc., are only raised to the first power.
- There are no products of \(y\) and its derivatives.
- \(a(x)y' + b(x)y = g(x)\)
- \(y' + \frac{3}{x} y = \frac{1}{x^3}\)
