Question

Test for exactness and solve the following equations. (a) $$ \left(\frac{1}{x}-\frac{1}{y}\right) d x+\frac{x}{y^{2}} d y=0 $$ (b) \((y \cos x-\sin y) d x+(\sin x-x \cos y) d y=0\) (c) \((1-x y)^{-2} d x+\left[y^{2}+x^{2}(1-x y)^{-2}\right] d y=0\)

Short answer

Question: Determine if the following differential equations are exact and, if so, solve them: a) (1/x - 1/y)dx + x/y^2 dy = 0 b) (ycos(x) - sin(y))dx + (sin(x) - xcos(y))dy = 0 c) (1-xy)^{-2}dx + (y^2 + x^2(1-xy)^{-2})dy = 0 Answer: a) The differential equation is exact, and the solution is -x/y = ln|x| + C b) The differential equation is not exact, so we cannot solve it using this method. c) The differential equation is exact, and the solution is -1/(y(1-xy)) + y^3/3 = C

Step-by-step solution

(a) Test for exactness

In this case, we have M(x, y) = (1/x - 1/y) and N(x, y) = x/y^2. Now we will calculate their partial derivatives: \(\frac{\partial M}{\partial y} = \frac{1}{y^2}\) and \(\frac{\partial N}{\partial x} = \frac{1}{y^2}\) Since ∂M/∂y = ∂N/∂x, the differential equation is exact.

(a) Solve the differential equation

Now that we know the equation is exact, we will integrate M with respect to x and N with respect to y: \(\int M dx = \int (1/x - 1/y) dx = \ln |x| - \frac{x}{y} + g(y)\) \(\int N dy = \int x/y^2 dy = -\frac{x}{y} + h(x)\) Equating the two results, we get: \(\ln |x| - \frac{x}{y} + g(y) = -\frac{x}{y} + h(x)\) By comparing the two sides, we can see that g(y) is a constant and h(x) = ln|x| + C, where C is an integration constant. Thus, the solution to the differential equation is: \(\boxed{-\frac{x}{y} = \ln |x| + C}\)

(b) Test for exactness

In this case, we have M(x, y) = ycos(x) - sin(y) and N(x, y) = sin(x) - xcos(y). Now we will calculate their partial derivatives: \(\frac{\partial M}{\partial y} = \cos x - \cos y\) and \(\frac{\partial N}{\partial x} = \cos x + y\sin y\) Since ∂M/∂y ≠ ∂N/∂x, the differential equation is not exact. Therefore, we cannot solve it using this method.

(c) Test for exactness

In this case, we have M(x, y) = (1-xy)^{-2} and N(x, y) = y^2 + x^2(1-xy)^{-2}. Now we will calculate their partial derivatives: \(\frac{\partial M}{\partial y} = 2xy(1-xy)^{-3}\) and \(\frac{\partial N}{\partial x} = 2x(1-xy)^{-2} + 2x^2y(1-xy)^{-3}\) Since ∂M/∂y = ∂N/∂x, the differential equation is exact.

(c) Solve the differential equation

Now that we know the equation is exact, we will integrate M with respect to x and N with respect to y: \(\int M dx = \int (1-xy)^{-2} dx = -\frac{1}{y(1-xy)} + g(y)\) \(\int N dy = \int [y^2 + x^2(1-xy)^{-2}] dy = y^3/3 - \frac{x^2}{y(1-xy)} + h(x)\) Equating the two results, we get: \(-\frac{1}{y(1-xy)} + g(y) = \frac{y^3}{3} - \frac{x^2}{y(1-xy)} + h(x)\) By comparing the two sides, we can see that g(y) = y^3/3 and h(x) is a constant. Thus, the solution to the differential equation is: \(\boxed{-\frac{1}{y(1-xy)} + \frac{y^3}{3} = C}\)

Key concepts

Test for Exactness

Determining whether a differential equation is exact is a fundamental step in the analytical method of solving such equations. An equation of the form \(M(x, y)dx + N(x, y)dy = 0\) is said to be exact if the partial derivatives of \(M\) with respect to \(y\) and \(N\) with respect to \(x\) are equal, that is \(\frac{\partial M}{\partial y} = \frac{\partial N}{\partial x}\).

To perform the test for exactness, compute these partial derivatives and compare them. If they match, you have an exact differential equation, which implies the existence of a function \(F(x, y)\) such that \(dF = Mdx + Ndy\). Notably, in the provided examples, part (a) and (c) are exact, while part (b) is not exact since the partial derivatives are not the same. This initial test plays a crucial role in defining the solving strategy.

Solve Differential Equations

Solving a differential equation, particularly an exact one, involves finding a function \(F(x, y)\) whose total differential matches the given equation. For an exact equation, this is accomplished by separately integrating \(M\) with respect to \(x\) and \(N\) with respect to \(y\), keeping in mind to add an arbitrary function of the other variable each time since the integration only gives us antiderivatives in terms of one variable.

The solutions from the partial integrations are then equated to solve for the arbitrary functions. When you solve for these functions correctly, you obtain the general solution \(F(x, y) = C\), where \(C\) is the constant of integration. This process is clearly illustrated in steps 2 and 5 for equations (a) and (c), leading to the complete solution for these exact differential equations.

Partial Derivatives

Partial derivatives are a crucial concept when working with multivariable functions, such as when testing for the exactness of differential equations. They represent how the function changes as each variable is varied while holding the others constant. In the context of exact differential equations, you calculate the partial derivative of \(M\) with respect to \(y\) (\(\frac{\partial M}{\partial y}\)) and of \(N\) with respect to \(x\) (\(\frac{\partial N}{\partial x}\)).

The equality of these partial derivatives is essential for the function \(F(x, y)\) to exist such that its differential is the equation in question. The computations shown in the step by step solution reveal how the respective partial derivatives are obtained and compared.

Integration of Functions

Integration is a fundamental operation in calculus, often referred to as finding the antiderivative or the area under a curve. When dealing with exact differential equations, integration is used to reconstruct the function \(F(x, y)\) from its differential. This involves integrating \(M\) with respect to \(x\) and \(N\) with respect to \(y\), usually resulting in separate terms plus an integration constant or an arbitrary function of the other variable.

The integration of functions explains why in the solutions for equations (a) and (c), \(g(y)\) and \(h(x)\) were introduced. Those arbitrary functions are determined through the comparison of integration results, allowing for the synthesis of the full solution to the equation. Integration, then, is not just about finding areas, but also about constructing solutions to differential equations from their infinitesimal relationships.