Find solution of differential equation(Y+ Y/X + Sin Y )dX + (X+ ln X+ X Cos Y)dY=0

To solve the differential equation \((Y + \frac{Y}{X} + \sin Y) dX + (X + \ln X + X \cos Y) dY = 0\), we can start by rewriting it in a more manageable form. This equation is a first-order ordinary differential equation, and we can analyze it using the method of exact equations or by finding an integrating factor if necessary.

Rearranging the Equation

First, let's express the equation in the standard form:

Let \(M(X, Y) = Y + \frac{Y}{X} + \sin Y\) and \(N(X, Y) = X + \ln X + X \cos Y\). The equation can then be written as:

\(M dX + N dY = 0\)

Checking for Exactness

To determine if this equation is exact, we need to check if the following condition holds:

\(\frac{\partial M}{\partial Y} = \frac{\partial N}{\partial X}\)

Calculating the partial derivatives:

• \(\frac{\partial M}{\partial Y} = 1 + \frac{1}{X} + \cos Y\)
• \(\frac{\partial N}{\partial X} = 1 + \frac{1}{X} + \cos Y\)

Since both derivatives are equal, the equation is indeed exact.

Finding the Potential Function

Next, we need to find a function \(F(X, Y)\) such that:

• \(\frac{\partial F}{\partial X} = M\)
• \(\frac{\partial F}{\partial Y} = N\)

Integrating M with Respect to X

We start by integrating \(M\) with respect to \(X\):

\(F(X, Y) = \int (Y + \frac{Y}{X} + \sin Y) dX = YX + Y \ln X + X \sin Y + g(Y)\)

Here, \(g(Y)\) is an arbitrary function of \(Y\) that arises because the integration is with respect to \(X\).

Finding g(Y)

Now, we differentiate \(F(X, Y)\) with respect to \(Y\) to find \(g(Y)\):

\(\frac{\partial F}{\partial Y} = X + \ln X + X \cos Y + g'(Y)\)

Setting this equal to \(N\):

\(X + \ln X + X \cos Y + g'(Y) = X + \ln X + X \cos Y\)

This implies that \(g'(Y) = 0\), meaning \(g(Y)\) is a constant. We can denote this constant as \(C\).

Final Form of the Solution

Thus, the potential function simplifies to:

\(F(X, Y) = YX + Y \ln X + X \sin Y = C\)

This represents the implicit solution to the original differential equation. To express it explicitly, you may need to isolate \(Y\) depending on the context or specific values of \(X\) and \(Y\).

Summary

In summary, we have solved the differential equation by recognizing it as exact, finding the potential function, and arriving at the implicit solution. The process involved checking for exactness, integrating the components, and determining the arbitrary function. This method is quite powerful for first-order differential equations that can be expressed in this form.