Question

Solve each differential equation by first finding an integrating factor. $$ \left(5 x y+4 y^{2}+1\right) d x+\left(x^{2}+2 x y\right) d y=0. $$

Short answer

The short answer to the given differential equation is as follows: First, we find the integrating factor (IF): \[ IF = \exp\left(\int \left(\frac{1}{x^2+2xy}\left(5x + 8y\right) - \frac{1}{5xy + 4y^2 + 1}\left(2x + 2y\right)\right)dx\right) \] Then, we multiply the given differential equation with the integrating factor \(u(x, y)\) and solve the exact differential equation to obtain the general solution: \[ F(x, y) = C, \] where \(C\) is an arbitrary constant.

Step-by-step solution

Identify the given differential equation and rearrange

The given differential equation is: \[ \left(5xy + 4y^2 + 1\right) dx + \left(x^2 + 2xy\right) dy = 0 \] Let's rewrite the equation as follows: \[ M(x, y)dx + N(x, y)dy = 0 \] where \(M(x, y) = 5xy + 4y^2 + 1\) and \(N(x, y) = x^2 + 2xy\).

Find integrating factor

To find the integrating factor (IF), we need to compute \(\frac{1}{N}\frac{\partial M}{\partial y} - \frac{1}{M}\frac{\partial N}{\partial x}\) and integrate it with respect to \(x\). First, let's find the partial derivatives \(\frac{\partial M}{\partial y}\) and \(\frac{\partial N}{\partial x}\): \[ \frac{\partial M}{\partial y} = 5x + 8y \] \[ \frac{\partial N}{\partial x} = 2x + 2y \] Compute \(\frac{1}{N}\frac{\partial M}{\partial y} - \frac{1}{M}\frac{\partial N}{\partial x}\): \[ \frac{1}{x^2+2xy}\left(5x + 8y\right) - \frac{1}{5xy + 4y^2 + 1}\left(2x + 2y\right) \] Now, integrate with respect to \(x\): \[ IF = \exp\left(\int \left(\frac{1}{x^2+2xy}\left(5x + 8y\right) - \frac{1}{5xy + 4y^2 + 1}\left(2x + 2y\right)\right)dx\right) \]

Multiply the given differential equation by integrating factor

Let's assume the integrating factor (IF) to be \(u(x, y)\): Now, multiply the given differential equation with the integrating factor \(u(x, y)\), we obtain: \[ u(x, y)(M(x, y)dx + N(x, y)dy) = 0 \] This differential equation is exact.

Solve the exact differential equation for the solution

Now, we need to solve the exact differential equation by finding the function \(F(x, y)\) such that \(\frac{\partial F}{\partial x} = u(x, y)M(x, y)\) and \(\frac{\partial F}{\partial y} = u(x, y)N(x, y)\). First, integrate \(\frac{\partial F}{\partial x} = u(x, y)M(x, y)\) with respect to \(x\), keeping constant of integration: \[ F(x, y) = \int u(x, y) M(x, y)dx + f(y) \] Now differentiate the above expression with respect to y: \[ \frac{\partial F}{\partial y} = \frac{\partial}{\partial y} \int u(x, y) M(x, y) dx + \frac{\partial f(y)}{\partial y} \] Since \(\frac{\partial F}{\partial y} = u(x, y)N(x, y)\), we have: \[ \frac{\partial}{\partial y} \int u(x, y) M(x, y) dx + \frac{\partial f(y)}{\partial y} = u(x, y)N(x, y) \] Now, solve this equation for \(f(y)\) and substitute it back into the expression for \(F(x, y)\). Then, the general solution to the given differential equation is given by: \[ F(x, y) = C \] where \(C\) is an arbitrary constant.

Key concepts

Exact Differential Equations

When solving ordinary differential equations (ODEs), one powerful technique involves the concept of exact differential equations. An exact differential equation is of the form:
\(M(x, y)dx + N(x, y)dy = 0\) where \(M\) and \(N\) have continuous partial derivatives. An equation is exact if
\(\frac{\partial M}{\partial y} = \frac{\partial N}{\partial x}\).
This condition arises from Clairaut's theorem on equality of mixed partial derivatives for a function \(F(x, y)\) such that \(M = \frac{\partial F}{\partial x}\) and \(N = \frac{\partial F}{\partial y}\).

• When the equation is not naturally exact, an integrating factor can be found to multiply the equation and make it exact.
• The integrating factor is typically a function of \(x\), \(y\), or both, determined by integrating a certain expression derived from \(M\) and \(N\).
• Once the equation is made exact using the integrating factor, the next step is to find a potential function \(F(x, y)\) whose partial derivatives equal \(M\) and \(N\) respectively.

The solution process involves integrating \(M\) with respect to \(x\) and \(N\) with respect to \(y\), including the arbitrary functions of integration, and then ensuring that they yield the same result for \(F(x, y)\).

Ordinary Differential Equations

Ordinary Differential Equations (ODEs) are equations involving derivatives of an unknown function with respect to a single variable. The order of an ODE is determined by the highest derivative present. In the given exercise, the differential equation is first-order as it involves first derivatives only.

Importance of ODEs

They are essential in various fields such as physics, engineering, and economics as they describe a multitude of dynamic phenomena. Finding solutions to ODEs is a fundamental aspect of mathematical modeling.

• A solution to an ODE represents a function that satisfies the equation.
• For first-order ODEs, the general solution will include an arbitrary constant, while higher-order ODEs will have as many constants as the order of the equation.
• The integrating factor method is one way to solve certain types of first-order ODEs by making them exact.

The goal is to simplify the ODE and find an explicit equation for the unknown function or describe the function implicitly in terms of its variables.

Partial Derivatives

Partial derivatives play a pivotal role in solving exact differential equations as they elucidate how a function changes as one variable varies while keeping the other variable fixed.

Computing Partial Derivatives

For a function \(f(x, y)\), the partial derivative with respect to \(x\) is denoted as \(\frac{\partial f}{\partial x}\) and provides the rate at which \(f\) changes when \(x\) changes, with \(y\) held constant. Likewise, the partial derivative with respect to \(y\), \(\frac{\partial f}{\partial y}\), indicates the rate of change of \(f\) with respect to \(y\).

• The concept of partial derivatives allows us to test whether a given ODE is exact.
• In the context of the exercise, the coefficients of \(dx\) and \(dy\) -- \(M(x, y)\) and \(N(x, y)\) -- are functions for which we compute partial derivatives to find the integrating factor.
• Partial derivatives are also employed to integrate the multiplied integrating factor into the ODE to solve for the potential function \(F(x, y)\).

The correct application of partial derivatives ensures that the combination of integrating factors with the original ODE creates a fully integrable equation that ultimately reveals the general solution.