Question

Solving a Homogeneous Differential Equation In Exercises \(75-80\) , solve the homogeneous differential equation in terms of \(x\) and \(y .\) A homogeneous differential equation is an equation of the form \(M(x, y) d x+N(x, y) d y=0,\) where \(M\) and \(N\) are homogeneous functions of the same degree. To solve an equation of this form by the method of separation of variables, use the substitutions \(y=v x\) and \(d y=x d v+v d x\) $$ (x-y) d x-(x+y) d y=0 $$

Short answer

The solution to the differential equation is \( y = Cx \), where C is the constant of integration.

Step-by-step solution

Identify homogeneous equation and Substitutions

Given equation is \((x-y) dx - (x+y) dy = 0\). This equation is homogeneous and the substitutions given are \(y= vx\) and \(dy= x dv + v dx\)

Substitute

Substitute \(y\) and \(dy\) in terms of \(v\) and \(x\) into the equation. So, the equation becomes \((x - vx)dx - (x+ vx)(x dv + v dx) = 0). After simplification, the equation is: dx - v dx - (x dv + v dx) - vx dv =0 \)

Simplifying

After rearrangement and simplification, on gathering the like terms, the equation gets simplified to dx - v dx - x dv - v dx - vx dv = 0 which simplifies to \( -x dv = 0\)

Separation of Variables

The equation \(-x dv = 0\) can be divided by \( -x\) on both sides to get \(dv = 0\)

Integrate

Integrating both sides, we get \( ∫ dv =∫0 dt\), which results in \(v = C\), where C is the constant of integration

Substitute v back to original

Now, substitute \( v = C \) back in terms of \( y \), which gives our solution as \( y = Cx \)

Key concepts

Separation of Variables

Separation of variables is a powerful technique often used to solve differential equations. It's particularly useful when dealing with a homogeneous differential equation, such as the one given: \((x-y)dx-(x+y)dy=0\).

This technique involves manipulating the equation to isolate each variable on a different side of the equation. In our specific exercise, we make use of the substitutions \(y = vx\) and \(dy = xdv + vdx\). These substitutions help reshape the equation into a form that allows us to separate the variables.

The goal is to express the equation such that all terms involving \(x\) are on one side and all terms involving \(y\) are on the other. By doing this, we can use integration to solve the equation.

Integration

Once we have separated the variables, we can proceed with integration. Integration is the process of finding the integral of a function, which is the reverse operation of differentiation.

In the exercise, after separation of variables, we end up with the equation \( dv = 0 \). We carry out integration on both sides of the equation:

• \( \int dv = \int 0 dt \)

Integrating the left side, we get \(v\), and the right side simply integrates to the constant of integration, \(C\).

This process completes the integration step, leading us to the next important component of solving the equation: the constant of integration.

Homogeneous Functions

The term 'homogeneous function' plays a significant role in solving differential equations like the one in the exercise. A function is considered homogeneous if it has terms that are of the same degree.

In our specific case, the given equation \((x-y)dx-(x+y)dy=0\) is a homogeneous equation because both \(M(x, y)\) and \(N(x, y)\) are functions of the same degree.

The importance of recognizing homogeneous functions lies in the suitability for using the substitutions \(y = vx\) and \(dy = x dv + v dx\). These substitutions allow us to reduce the complexity of the equation and effectively apply separation of variables to advance the solution.

Constant of Integration

The constant of integration, often represented as \(C\), emerges during the process of integration. It represents an arbitrary constant that arises because integrating a derivative only gives us one of possibly many functions.

In our exercise, after integrating \( dv = 0 \), we find \( v = C \). This constant is crucial because it encompasses all possible solutions that satisfy the differential equation.

When we convert back to the original variables by substituting \( v = C \) into \( y = vx \), we find the general solution \( y = Cx \). The constant of integration represents a family of straight lines (since \( C \) can take on any real value) that are solutions to the given differential equation.