Solving a Differential Equation with Variable Substitution

Beyond the basics of calculus, solving differential equations often involves creative problem-solving, including the use of variable substitution. In this article, we will walk through the process of solving a specific differential equation using such a technique. The goal is to highlight how to manipulate and simplify complex equations to make them more tractable.

Problem Setup

Consider the differential equation:

y cosleft(frac{y}{x}right) x sinleft(frac{y}{x}right) cosleft(frac{y}{x}right) frac{dy}{dx}

To solve this, we will use a variable substitution method, breaking it down step-by-step.

Step 1: Substitution

Let us set:

y t x

and derive:

frac{dy}{dx} t x frac{dt}{dx}

Substituting (y) and (frac{dy}{dx}) into the original equation, we get:

t x cosleft(frac{t x}{x}right) x sinleft(frac{t x}{x}right) cosleft(frac{t x}{x}right) (t x frac{dt}{dx})

Simplify the trigonometric terms:

t x cos t x sin t cos t (t x frac{dt}{dx})

After simplification, we have:

x cos t (t cos t - sin t) x cos t (frac{dt}{dx})

Dividing both sides by (x cos t), we get:

t cos t - sin t frac{dt}{dx}

Step 2: Further Substitution and Rearrangement

To further simplify, let's make a new substitution:

u frac{y}{x}

This means:

y x u

And:

frac{dy}{dx} u x frac{du}{dx}

Substitute into the simplified equation:

x u cos u (u cos u - sin u) x u cos u (u x frac{du}{dx})

Notice that (x u cos u eq 0), we can divide both sides by (x u cos u):

u cos u - sin u u x frac{du}{dx}

Rearrange to get a separable differential equation:

x frac{du}{dx} -u cos u - sin u

Step 3: Separating Variables

Separate the variables by dividing both sides by (u cos u sin u):

frac{dx}{x} -frac{du}{u cos u sin u}

Integrate both sides:

int frac{dx}{x} -int frac{du}{u cos u sin u}

The left side is:

ln|x| -int frac{du}{u cos u sin u}

For the right side, we recognize that a known integral solution does not exist in elementary functions. We will denote the integral as:

I int frac{du}{u cos u sin u}

Thus the solution is:

x e^x A e^{-I}

Where (A) is a constant determined by initial conditions.

Final Step: Back-Substitution

Returning to the original variable (y), we have:

y x A x^2 e^{-I}

Conclusion

In summary, we have successfully transformed a complex differential equation into a more manageable form and arrived at its final integral solution. This method showcases the power of variable substitution and integration in solving challenging differential equations. The process includes careful manipulation, substitution, and integration techniques to simplify the equation and find a general solution.

Keywords: differential equation, variable substitution, integral solution