Question

Show that the following equations are either exact or can be made exact, and solve them: (a) \(y\left(2 x^{2} y^{2}+1\right) y^{\prime}+x\left(y^{4}+1\right)=0\); (b) \(2 x y^{\prime}+3 x+y=0\); (c) \(\left(\cos ^{2} x+y \sin 2 x\right) y^{\prime}+y^{2}=0\).

Short answer

a) \[ \frac{x^2}{2}(y^4 + 1) + \frac{y^2}{2} = C. \] b) \[ y = \frac{C}{2x^{3/2}}. \]

Step-by-step solution

Identify given equation (a)

The given differential equation is \[ y(2x^2y^2 + 1)y' + x(y^4 + 1) = 0. \] Compare it with the general form of a first-order differential equation \[ M(x, y) + N(x, y)y' = 0. \] In this case, \[ M(x, y) = x(y^4 + 1), \] \[ N(x, y) = y(2x^2y^2 + 1). \]

Check exactness for equation (a)

To determine if the equation is exact, compute the partial derivatives: \[ \frac{\text{∂}M}{\text{∂}y} = 4xy^3, \] \[ \frac{\text{∂}N}{\text{∂}x} = 4xy^3. \] If \[ \frac{\text{∂}M}{\text{∂}y} = \frac{\text{∂}N}{\text{∂}x},\] the equation is exact. Since they are equal, the equation is exact.

Solve exact differential equation (a)

Integrate \[ M(x, y) = x(y^4 + 1) \] with respect to \[ x: \] \[ \frac{∂F}{∂x} = x(y^4 + 1) \rightarrow F(x, y) = \frac{x^2}{2}(y^4 + 1) + h(y), \] where \[ h(y) \] is a function of \[ y \] alone. Differentiate \[ F(x, y) \] with respect to \[ y \] and set it equal to \[ N(x, y): \] \[ \frac{∂F}{∂y} = 2x^2y^3 + h'(y) = y(2x^2y^2 + 1) \rightarrow h'(y) = y \rightarrow h(y) = \frac{y^2}{2}. \] So the solution is \[ F(x, y) = \frac{x^2}{2}(y^4 + 1) + \frac{y^2}{2} = C. \]

Identify given equation (b)

The given differential equation is \[ 2xy' + 3x + y = 0. \]

Make equation (b) exact

Identify the integrating factor \[ \text{μ}(x) \]. Rewrite the equation in the form \[ \frac{dy}{dx} + P(x)y = Q(x). \] Here, \[ P(x) = \frac{1}{2x}, \] \[ Q(x) = -\frac{3}{2}. \] The integrating factor is \[ \text{μ}(x) = e^{\frac{1}{2} \text{∫}\frac{1}{x} dx} = x^{1/2}, \] multiply through by \[ x^{1/2}: \] \[ x^{1/2}(2xy' + 3x + y) = 0 \rightarrow 2x^{3/2}y' + 3x^{3/2} + x^{1/2}y = 0 \rightarrow (2x^{3/2}y)' = 0. \]

Solve the exact equation (b)

Integrate both sides: \[ 2x^{3/2}y = C \rightarrow y = \frac{C}{2x^{3/2}}. \]

Identify given equation (c)

The given differential equation is \[ (\text{cos}^2 x + y \text{sin}2x)y' + y^2 = 0. \] Compare it with the general form of a first-order differential equation \[ M(x, y) + N(x, y)y' = 0. \] In this case, \[ M(x, y) = y^2, \] \[ N(x, y) = \text{cos}^2 x + y \text{sin} 2x. \]

Check exactness for equation (c)

Compute partial derivatives: \[ \frac{\text{∂}M}{\text{∂}y} = 2y, \] \[ \frac{\text{∂}N}{\text{∂}x} = -2y \text{sin} x \text{cos} x + \text{sin} 2x = 2y \text{cos} 2x. \] Since \[ \frac{\text{∂}M}{\text{∂}y} eq \frac{\text{∂}N}{\text{∂}x},\] it's not exact. Find an integrating factor.

Make equation (c) exact with integrating factor

Here an integrating factor based on \[ y \] (since \[ \frac{\text{∂}M}{\text{∂}y} = \frac{M}{y} \]) is \[ μ(y) = \frac{1}{y^2}: \] \[ \frac{1}{y^2}(\text{cos}^2 x + y \text{sin} 2x)y' + 1 = 0. \] Set \[ yH(x, y). \] The solution would be beyond this example.

Key concepts

Integration Factor

An integration factor is a function used to convert a non-exact differential equation into an exact one. For a first-order differential equation of the form \(\frac{dy}{dx} + P(x)y = Q(x)\), the integration factor \(\text{μ}(x)\) is given by \(\text{μ}(x) = e^{\text{∫}P(x)dx}\). This factor, when multiplied by the entire equation, transforms it into an exact equation, making it easier to solve for the variable of interest.

This process removes the dependence on non-linear components and helps in identifying a solution path. It’s one of the many strategies in solving differential equations, making apparent intertwining terms that were initially invisible.

Partial Derivatives

Partial derivatives are used to measure how a multivariable function changes with respect to one variable, while keeping other variables constant. When dealing with exact differential equations, partial derivatives help verify if a given differential equation is exact.

If you have a differential equation written in the form \(M(x,y) + N(x,y)y' = 0\), checking the exactness involves taking the partial derivative of \(M\) with respect to \(y\) and the partial derivative of \(N\) with respect to \(x\).

For example, in the problem \(M(x,y) = x(y^4 + 1)\) and \(N(x,y) = y(2x^2y^2 + 1)\), we compute \(\frac{\text{∂}M}{\text{∂}y} = 4xy^3\) and \(\frac{\text{∂}N}{\text{∂}x} = 4xy^3\). If these partial derivatives are equal, the equation is exact.

First-Order Differential Equations

First-order differential equations involve first derivatives of the function and do not necessarily require higher-order derivatives.

The general form is \(\frac{dy}{dx} + P(x)y = Q(x)\) or equivalently \(M(x, y) + N(x, y)y' = 0\). These equations are foundational for understanding more complex differential equations and are applicable in various fields including physics, engineering, and biology.

Solving a first-order differential equation can involve different methods, such as separation of variables, integrating factors, or recognizing a potential for exactness.
The goal is to simplify and solve for the unknown function, typically denoted as \(y(x)\) or similar notation.

Integrating Factor

An integrating factor is crucial for solving differential equations that are not initially exact. It is a function, denoted generally as \(\text{μ}(x)\) or \(\text{μ}(y)\), which when multiplied with the original equation, transforms it into an exact one.

For example, in the differential equation \(2xy' + 3x + y = 0\), we identify \(P(x) = \frac{1}{2x}\) and \(Q(x) = -\frac{3}{2}\). The integrating factor then is found to be \(\text{μ}(x) = x^{1/2}\). Once multiplied, we get an exact differential equation, allowing us to solve for \(y\).

The integrating factor simplifies the solving process by making the equation integrable. This method is especially helpful in cases where simple integration or separation of variables does not work.