Chapter 2: Problem 3
Solve each differential equation by first finding an integrating factor. \(\left[y^{2}(x+1)+y\right] d x+(2 x y+1) d y=0\)
Short Answer
Expert verified
The general solution for the given differential equation using the integrating factor is:
\((2xy+1)y = -\int y^2(x+1)\,dx + C\)
Step by step solution
01
Rewrite the equation into standard form
We need to rewrite the given differential equation as: \(\frac{dy}{dx} + P(x)y = Q(x)\). So, divide both sides by \(dx\cdot dy\) to get:
\(\frac{dy}{dx} = -\frac{y^{2}(x+1) + y}{2xy + 1}\)
This gives us the standard form: \(\frac{dy}{dx} + \frac{1}{2xy+1}y = -\frac{y^2(x+1)}{2xy+1}\).
02
Find the integrating factor
The integrating factor is given by:
\(IF(x) = e^{\int P(x) dx} = e^{\int \frac{1}{2xy+1} dx}\)
Since we cannot solve this integral directly, we will make the substitution \(u = 2xy + 1\), yielding \(du = 2y\, dx\). Now, our integral simplifies to:
\(IF(x) = e^{\int \frac{1}{u} du} = e^{\ln |u|} = u\)
Putting the original expression for \(u\), we have the integrating factor:
\(IF(x) = 2xy + 1\)
03
Multiply by the integrating factor
Now, multiply the entire differential equation by the integrating factor:
\((2xy + 1)\left(\frac{dy}{dx} + \frac{1}{2xy+1}y\right) = -(2xy+1) \frac{y^2(x+1)}{2xy+1}\)
After multiplying, it simplifies to:
\(\frac{d}{dx}[(2xy + 1)y] = - y^2(x+1)\)
04
Integrate both sides
Now, integrate both sides with respect to x:
\(\int\frac{d}{dx}[(2xy+1)y]\,dx = -\int y^2(x+1)\,dx\)
Integrating the left side, we have:
\((2xy+1)y = -\int y^2(x+1)\,dx + C\) (where C is the constant of integration)
The integral on the right side is not trivial, so we will leave it as-is and write our solution:
\((2xy+1)y = -\int y^2(x+1)\,dx + C\)
This is the general solution for the given differential equation using the integrating factor.
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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 clever tool used in solving first-order linear differential equations. It transforms a non-exact equation into an exact one, making it easier to solve. To use the integrating factor technique, you must first express the differential equation in standard linear form: \(\frac{dy}{dx} + P(x)y = Q(x)\). Here, \(P(x)\) and \(Q(x)\) are functions of \(x\). The integrating factor, often denoted as \(\mu(x)\), is found using the formula:
- \(\mu(x) = e^{\int P(x) dx}\)
- \(\mu(x)\frac{dy}{dx} + \mu(x)P(x)y = \mu(x)Q(x)\)
First-order Differential Equations
First-order differential equations involve derivatives of a function with respect only to one variable. These equations include the first derivative, \(\frac{dy}{dx}\), being the highest order derivative present. They are foundational in describing various natural phenomena, from population growth to physical laws. The simplest form of first-order differential equations is separable equations, expressed as:
Another important form is the linear first-order differential equation given by:
- \(h(y) dy = g(x) dx\)
Another important form is the linear first-order differential equation given by:
- \(\frac{dy}{dx} + P(x)y = Q(x)\)
Solution Techniques for Differential Equations
Solving differential equations requires various techniques depending on the form and complexity of the equation. For first-order differential equations, three primary techniques include separation of variables, integrating factors, and exact equations.For equations in the form \(\frac{dy}{dx} + P(x)y = Q(x)\), utilizing an integrating factor simplifies finding a solution. The integrating factor makes the left-hand side a derivative, allowing straightforward integration.
- **Separation of Variables:** This technique is ideal for separable differential equations where variables can be separated on opposite sides of the equation. It's solved by integrating both sides.
- **Integrating Factor:** As explained, this is used for non-separable first-order linear differential equations. The method converts the equation to an exact one.
- **Exact Equations:** If an equation is already exact, it's typically in the form \(M(x, y) dx + N(x, y) dy = 0\), where \(\frac{\partial M}{\partial y} = \frac{\partial N}{\partial x}\). Solving involves finding a potential function, which simplifies the equation to its implicit solution.