Solving 2y Sin 2x Dx - (1 + Y² + Cos² X) Dy = 0 A Step-by-Step Guide

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This article delves into the solution of the differential equation 2y sin 2x dx - (1 + y² + cos² x) dy = 0. We will explore the methods required to solve this equation, providing a step-by-step guide and highlighting key concepts in differential equations. This comprehensive explanation will benefit students, educators, and anyone interested in understanding the techniques for solving such problems. By the end of this article, you will have a clear understanding of how to approach and solve this particular differential equation, as well as a broader grasp of the principles involved in solving similar equations.

1. Understanding the Differential Equation

In this section, we will focus on understanding the given differential equation. The equation presented is 2y sin 2x dx - (1 + y² + cos² x) dy = 0. Before diving into the solution, it is crucial to identify the type of differential equation we are dealing with. This identification guides the selection of the appropriate solution method. To begin, let’s rearrange the equation to a more standard form.

We can rewrite the equation as:

(2y sin 2x) dx = (1 + y² + cos² x) dy

Now, let's divide both sides by dx and (1 + y² + cos² x) to separate the variables, if possible. This gives us:

(2y sin 2x) / (1 + y² + cos² x) = dy/dx

From this form, it’s not immediately clear that the variables can be easily separated. This suggests that the equation might not be a simple separable differential equation. Another approach is to check if the equation is exact. An exact differential equation is one that can be written in the form:

M(x, y) dx + N(x, y) dy = 0

where ∂M/∂y = ∂N/∂x. In our case, we have:

M(x, y) = 2y sin 2x

N(x, y) = -(1 + y² + cos² x)

Let's compute the partial derivatives:

∂M/∂y = ∂(2y sin 2x)/∂y = 2 sin 2x

∂N/∂x = ∂(-(1 + y² + cos² x))/∂x = -∂(cos² x)/∂x

To find the derivative of cos² x, we use the chain rule:

∂(cos² x)/∂x = 2 cos x * (-sin x) = -2 sin x cos x = -sin 2x

So, ∂N/∂x = -(-sin 2x) = sin 2x

Comparing the partial derivatives, we see that ∂M/∂y = 2 sin 2x and ∂N/∂x = sin 2x. Since these are not equal, the given differential equation is not exact in its current form. However, it might be possible to make it exact by finding an integrating factor. An integrating factor is a function that, when multiplied by the differential equation, makes it exact. Before we try to find an integrating factor, let’s explore other methods to ensure we choose the most efficient approach.

Another perspective is to examine if the equation is homogeneous. A homogeneous differential equation can be written in the form dy/dx = f(y/x). Our equation doesn't immediately fit this form. However, we should explore further to make a definitive conclusion.

Considering all these aspects, we have determined that the differential equation is not separable and not exact in its current form. The next step is to investigate potential integrating factors or consider other advanced methods. We'll delve into these methods in the subsequent sections to effectively solve the given differential equation.

2. Checking for Exactness and Finding the Integrating Factor

In the previous section, we determined that the differential equation 2y sin 2x dx - (1 + y² + cos² x) dy = 0 is not exact in its original form. To solve this type of equation, we often seek an integrating factor. This is a function, say μ(x, y), that we multiply through the equation to make it exact. An exact equation is one where the partial derivative of the coefficient of dx with respect to y equals the partial derivative of the coefficient of dy with respect to x.

Recall that our equation is in the form:

M(x, y) dx + N(x, y) dy = 0

Where:

M(x, y) = 2y sin 2x

N(x, y) = -(1 + y² + cos² x)

We calculated the partial derivatives as:

∂M/∂y = 2 sin 2x

∂N/∂x = sin 2x

Since ∂M/∂y ≠ ∂N/∂x, the equation is not exact. To find an integrating factor, we can use the following formulas:

  1. If (∂M/∂y - ∂N/∂x) / N is a function of x only, say f(x), then the integrating factor μ(x) is given by:

    μ(x) = e^(∫ f(x) dx)

  2. If (∂N/∂x - ∂M/∂y) / M is a function of y only, say g(y), then the integrating factor μ(y) is given by:

    μ(y) = e^(∫ g(y) dy)

Let's calculate (∂M/∂y - ∂N/∂x) / N:

(∂M/∂y - ∂N/∂x) / N = (2 sin 2x - sin 2x) / -(1 + y² + cos² x) = sin 2x / -(1 + y² + cos² x)

This expression is not a function of x alone because it contains y terms. Now let's calculate (∂N/∂x - ∂M/∂y) / M:

(∂N/∂x - ∂M/∂y) / M = (sin 2x - 2 sin 2x) / (2y sin 2x) = -sin 2x / (2y sin 2x) = -1 / (2y)

This expression is a function of y only, so we can use the second formula to find the integrating factor μ(y):

μ(y) = e^(∫ g(y) dy) = e^(∫ (-1 / (2y)) dy) = e^(-1/2 ∫ (1/y) dy)

Integrating 1/y with respect to y gives ln|y|, so:

μ(y) = e^(-1/2 ln|y|) = e(ln|y|(-1/2)) = |y|^(-1/2)

For simplicity, we can take μ(y) = y^(-1/2) = 1/√y. Now we multiply the original differential equation by this integrating factor:

(2y sin 2x / √y) dx - ((1 + y² + cos² x) / √y) dy = 0

Simplifying, we get:

2√y sin 2x dx - (1/√y + y^(3/2) + cos² x / √y) dy = 0

Now, let's redefine M and N with the integrating factor applied:

M(x, y) = 2√y sin 2x

N(x, y) = -(1/√y + y^(3/2) + cos² x / √y)

We need to check if this new equation is exact. Compute the partial derivatives again:

∂M/∂y = ∂(2√y sin 2x)/∂y = 2 sin 2x * (1 / (2√y)) = sin 2x / √y

∂N/∂x = ∂(-(1/√y + y^(3/2) + cos² x / √y))/∂x = -(∂(cos² x / √y)/∂x) = -(-2 cos x sin x / √y) = sin 2x / √y

Now we have ∂M/∂y = ∂N/∂x, so the equation is exact after applying the integrating factor. In the next section, we will solve this exact differential equation.

3. Solving the Exact Differential Equation

Having found the integrating factor μ(y) = 1/√y and multiplied it through the original differential equation, we now have an exact differential equation. The equation is:

2√y sin 2x dx - (1/√y + y^(3/2) + cos² x / √y) dy = 0

As we verified in the previous section, this equation satisfies the condition for exactness, i.e., ∂M/∂y = ∂N/∂x, where:

M(x, y) = 2√y sin 2x

N(x, y) = -(1/√y + y^(3/2) + cos² x / √y)

To solve an exact differential equation, we need to find a function F(x, y) such that:

∂F/∂x = M(x, y)

∂F/∂y = N(x, y)

First, let's integrate M(x, y) with respect to x:

F(x, y) = ∫ M(x, y) dx = ∫ 2√y sin 2x dx

Treating y as a constant, we get:

F(x, y) = 2√y ∫ sin 2x dx = 2√y (-1/2 cos 2x) + g(y) = -√y cos 2x + g(y)

Here, g(y) is an arbitrary function of y, which arises because we are performing a partial integration with respect to x. Next, we need to find g(y). To do this, we differentiate F(x, y) with respect to y and set it equal to N(x, y):

∂F/∂y = ∂(-√y cos 2x + g(y))/∂y = -cos 2x * (1 / (2√y)) + g'(y)

Now, set ∂F/∂y equal to N(x, y):

-cos 2x / (2√y) + g'(y) = -(1/√y + y^(3/2) + cos² x / √y)

Isolate g'(y):

g'(y) = -1/√y - y^(3/2) - cos² x / √y + cos 2x / (2√y)

Recall the trigonometric identity cos 2x = 2 cos² x - 1. Substitute this into the equation:

g'(y) = -1/√y - y^(3/2) - cos² x / √y + (2 cos² x - 1) / (2√y)

g'(y) = -1/√y - y^(3/2) - cos² x / √y + cos² x / √y - 1 / (2√y)

g'(y) = -1/√y - y^(3/2) - 1 / (2√y)

Now, integrate g'(y) with respect to y to find g(y):

g(y) = ∫ g'(y) dy = ∫ (-1/√y - y^(3/2) - 1 / (2√y)) dy

g(y) = -∫ y^(-1/2) dy - ∫ y^(3/2) dy - 1/2 ∫ y^(-1/2) dy

g(y) = -2y^(1/2) - (2/5)y^(5/2) - y^(1/2) + C

g(y) = -3√y - (2/5)y^(5/2) + C

Substitute g(y) back into the expression for F(x, y):

F(x, y) = -√y cos 2x - 3√y - (2/5)y^(5/2) + C

The general solution to the exact differential equation is given by F(x, y) = constant. Therefore,

-√y cos 2x - 3√y - (2/5)y^(5/2) = C

This equation represents the implicit solution to the given differential equation. We have successfully navigated through the process of identifying the type of differential equation, finding an integrating factor, and solving the resulting exact equation. This step-by-step solution provides a clear understanding of the techniques involved.

4. Final Solution and Conclusion

In the previous sections, we meticulously worked through the steps to solve the differential equation 2y sin 2x dx - (1 + y² + cos² x) dy = 0. We identified that the equation was not exact in its original form, found an integrating factor μ(y) = 1/√y, and multiplied the equation by this factor to make it exact. We then solved the resulting exact differential equation by finding a function F(x, y) whose partial derivatives matched the coefficients of dx and dy. The final implicit solution we obtained is:

-√y cos 2x - 3√y - (2/5)y^(5/2) = C

where C is the constant of integration. This solution represents a family of curves in the xy-plane, each corresponding to a different value of C.

To summarize the process, we:

  1. Identified the differential equation as potentially non-exact.
  2. Calculated the partial derivatives ∂M/∂y and ∂N/∂x and confirmed that the equation was not exact.
  3. Found the integrating factor μ(y) = 1/√y using the formula μ(y) = e^(∫ g(y) dy), where g(y) = (∂N/∂x - ∂M/∂y) / M.
  4. Multiplied the original equation by the integrating factor to obtain an exact differential equation.
  5. Solved the exact equation by integrating M(x, y) with respect to x and then finding the function g(y) by differentiating F(x, y) with respect to y and comparing it with N(x, y).
  6. Obtained the general implicit solution F(x, y) = C.

This step-by-step approach is crucial for solving differential equations of this type. Understanding the underlying principles and techniques allows us to tackle similar problems effectively. The use of integrating factors is a powerful tool for transforming non-exact equations into exact ones, which can then be solved using standard methods. Furthermore, it showcases the importance of careful computation and the application of relevant calculus and trigonometric identities to simplify the expressions and arrive at the correct solution.

In conclusion, we have successfully solved the given differential equation by employing the method of integrating factors and exact differential equations. This detailed solution serves as a valuable resource for anyone studying or working with differential equations. The implicit solution -√y cos 2x - 3√y - (2/5)y^(5/2) = C provides a comprehensive understanding of the relationship between x and y that satisfies the original differential equation.