Solving the Differential Equation dy/dx y^2 sin 2x y sin x: A Comprehensive Guide

In this article, we will take a detailed look at solving the differential equation ( frac{dy}{dx} y^2 sin 2x y sin x ). We will explore various methods and techniques to approach this problem, including separation of variables and the Bernoulli equation method. Let's dive in!

Step 1: Rearranging the Equation

To start, we rearrange the given differential equation into a more standard form:

[ frac{dy}{dx} y^2 sin 2x y sin x ]

Step 2: Separation of Variables

Let's try to separate variables. First, we rewrite the equation in a form that allows us to separate ( y ) and ( x ):

[ frac{dy}{y sin 2x sin x} dx ]

Step 3: Integrating Both Sides

Now, we need to integrate both sides of the equation. The left-hand side can be tricky due to the presence of ( y ). We can rewrite it for easier integration:

[ int frac{dy}{y sin 2x sin x} int dx ]

However, this integral might require partial fraction decomposition or substitution depending on the form of ( y ).

Step 4: Solving the Integral

Assuming ( sin 2x - sin x eq 0 ), we proceed with integration:

[ int frac{1}{y} dy - int frac{sin x}{y sin 2x - sin x} dy x C ]

The first integral yields ( ln|y| ). The second integral can be more complicated but the goal is to isolate ( y ).

Step 5: Finding the General Solution

This equation is complex and the solution may not be expressible in elementary functions. Therefore, depending on the context or initial conditions, we can derive specific solutions. If you have initial conditions or specific values for ( x ) and ( y ), we can proceed with numerical methods or specific substitutions.

Alternative Approach: Using the Bernoulli Equation Method

The given equation is also a Bernoulli's type differential equation. In a Bernoulli equation, we divide the entire equation by ( y^{n-1} ) where ( n ) is the largest power of ( y ) that can be found in the equation. In this case, ( n 2 ), so we divide the whole equation by ( y^2 ):

[ frac{1}{y^2} frac{dy}{dx} frac{1}{y} sin x sin 2x ]

Substitution

We make the substitution ( t y^{1-n} ), i.e., ( t y^{-1} ) and ( frac{dt}{dx} -frac{1}{y^2} frac{dy}{dx} ):

[ -frac{dt}{dx} sin x sin 2x ]

Furthermore, we can rewrite the equation as:

[ frac{dt}{dx} - t sin x -sin 2x ]

Solving the Bernoulli Equation

This is now a first-order linear ordinary differential equation. We solve this by multiplying the whole equation with an integrating factor, which will make the left side of the equation equal to the derivative of a product:

[ mu(x) e^{int -sin x , dx} e^{cos x} ]

Then, the equation becomes:

[ e^{cos x} frac{dt}{dx} e^{cos x} t sin x -sin 2x e^{cos x} ]

Using the product rule, the left side simplifies to:

[ frac{d}{dx} (te^{cos x}) -sin 2x e^{cos x} ]

Now, we integrate both sides:

[ int frac{d}{dx} (te^{cos x}) , dx -int sin 2x e^{cos x} , dx ]

Using integration by parts, let:

[ u -cos x, , du sin x , dx ]

[ dv e^{cos x} sin x , dx, , v e^{cos x} ]

The integral on the right side becomes:

[ -int sin 2x e^{cos x} , dx -left( -cos x e^{cos x} e^{cos x} int e^{cos x} sin x , dx right) ]

Solving this, we get:

[ te^{cos x} 2e^{cos x}(cos x - 1) C ]

Thus, ( t 2(cos x - 1) e^{-cos x} C e^{-cos x} ) and:

[ frac{1}{y} 2(cos x - 1) e^{-cos x} C e^{-cos x} ]

Therefore, the general solution is:

[ y frac{e^{cos x}}{2(cos x - 1) C} ]

Summary

The given differential equation is separable and can be solved using various methods. The solution process involves careful integration and possibly numerical methods or special functions depending on the complexity of the integrals involved. If you have more specific requirements or initial conditions, please let me know!