Question

Identify each of the differential equations as to type (for example, separable, linear first order, linear second order, etc.), and then solve it. \((2 x-y \sin 2 x) d x+\left(2 y-\sin ^{2} x\right) d y=0\)

Short answer

The equation is exact. The solution is given by \[x^2 + y \cos 2x + y^2 = C.\]

Step-by-step solution

Identify the Type of Differential Equation

Rewrite the given differential equation as \[(2 x - y \sin 2 x) \, dx + (2 y - \sin^2 x) \, dy = 0.\]To classify the type, compare it to standard forms. Here, it doesn't fit separable, linear first-order, or linear second-order forms. Notice it's approach to a potential exact differential equation.

Check Exactness

To check for an exact differential equation, we need\[M(x, y) = 2x - y \sin 2x\]and\[N(x, y) = 2y - \sin^2 x.\]Calculate partial derivatives \[\frac{\partial M}{\partial y} = -\sin 2x\]and\[\frac{\partial N}{\partial x} = -2 \sin x \cos x = -\sin 2x.\]Since \frac{\partial M}{\partial y} = \frac{\partial N}{\partial x}, the equation is exact.

Solve the Exact Differential Equation

Find the function \Psi(x, y) such that \frac{\partial \Psi}{\partial x} = M and \frac{\partial \Psi}{\partial y} = N. Integrate \frac{\partial \Psi}{\partial x} w.r.t x,\[\Psi(x, y) = \int (2x - y \sin 2x) \, dx\]\[\Psi(x, y) = x^2 + y \cos 2x + g(y)\]Compute \frac{\partial \Psi}{\partial y}:\[\Psi_y = \cos 2x + \frac{d g(y)}{d y} = 2y - \sin^2 x.\]So \frac{d g(y)}{d y} = 2y. Integrate w.r.t y,\[g(y) = y^2 + C.\]

Form the General Solution

Combine everything to \[\Psi(x, y) = x^2 + y \cos 2x + y^2 = C.\]This is the implicit solution.

Key concepts

differential equations

Differential equations are equations that involve an unknown function and its derivatives. They can model many physical systems and processes, such as motion, heat, or waves. There are different types of differential equations:

• Separable: Can be rewritten as a product of two functions, each depending on a single variable.
• Linear First Order: In the form \(a(x) y' + b(x) y = c(x)\).
• Linear Second Order: In the form \(a(x) y'' + b(x) y' + c(x) y = d(x)\).

This particular problem involves identifying the type first, and determining if it fits an exact differential equation by checking if the exactness condition holds.

exactness condition

The exactness condition is essential for identifying exact differential equations. If a differential equation is in the form
\[ M(x, y) \, dx + N(x, y) \, dy = 0, \]
it is exact if the mixed partial derivatives of \(M\) and \(N\) are equal. Specifically, this means:
\[ \frac{\partial M}{\partial y} = \frac{\partial N}{\partial x}. \]
In our exercise:
\( M(x, y) = 2x - y \sin 2x \)
\( N(x, y) = 2y - \sin^2 x \)
we compute:
\( \frac{\partial M}{\partial y} = -\sin 2x \)
\( \frac{\partial N}{\partial x} = -2 \sin x \cos x = -\sin 2x \)
The derivatives are equal, confirming the equation is exact.

partial derivatives

Partial derivatives are derivatives of functions with more than one variable. They measure how the function changes as each variable changes while the others are held constant. For a function \(f(x, y)\):

• \( \frac{\partial f}{\partial x} \): Keeps \(y\) constant and differentiates with respect to \(x\).
• \( \frac{\partial f}{\partial y} \): Keeps \(x\) constant and differentiates with respect to \(y\).

In our problem, we used partial derivatives to check the exactness condition:
\( \frac{\partial M}{\partial y} = -\sin 2x \)
\( \frac{\partial N}{\partial x} = -\sin 2x \)
Both matched, allowing us to confirm the exactness of the equation.

integration

Integration is the process of finding the integral of a function, which is the reverse operation of differentiation. In the context of exact differential equations, solving involves finding a potential function \(\Psi(x, y)\) such that:
\(\frac{\partial \Psi}{\partial x} = M \)
\(\frac{\partial \Psi}{\partial y} = N \)
In our solution:
\(\frac{\partial \Psi}{\partial x} = 2x - y \sin 2x \)
Integrating w.r.t \(x\):
\(\Psi(x, y) = \int (2x - y \sin 2x) \, dx = x^2 + y \cos 2x + g(y) \)
Next, matching \(\frac{d g(y)}{\partial y}\) with \(N\):
\(\Psi_y = \cos 2x + \frac{d g(y)}{d y} = 2y - \sin^2 x \)
We solve for \(g(y)\):
\(\frac{d g(y)}{d y} = 2y \Rightarrow g(y) = y^2 \)
Thus the general solution is:
\(\Psi(x, y) = x^2 + y \cos 2x + y^2 = C \).