Question

$$ (x+4)\left(y^{2}+1\right) d x+y\left(x^{2}+3 x+2\right) d y=0 $$

Short answer

The general solution of the given differential equation is: \[\frac{y}{x} = - \int \frac{(x + 4)(v^2x^2 + 1) + 2v^2x(x^2 + 3x + 2)}{2v(x^2 + 3x + 2)} dx + C\]

Step-by-step solution

Verify the exactness of the equation

In order to determine if an equation is exact, we need to check if: \[ \frac{\partial M}{\partial y} = \frac{\partial N}{\partial x} \] We are given: \[ M(x, y) = (x + 4) (y^2 +1) \textrm{ and } N(x, y) = y (x^2 + 3x + 2) \] Calculate \(\frac{\partial M}{\partial y}\) and \(\frac{\partial N}{\partial x}\): \[ \frac{\partial M}{\partial y} = \frac{\partial}{\partial y}(x + 4) (y^2 +1) = (x + 4) \cdot 2y \] \[ \frac{\partial N}{\partial x} = \frac{\partial}{\partial x} y (x^2 + 3x + 2) = y(2x + 3) \] The equation is not exact, but we can notice that if we multiply N by 2: \[ 2\frac{\partial N}{\partial x} = 2y(2x + 3) \] Now, we verify the exactness of the equation: \[ (x+4)2y = 2y(2x+3) \] Since the \(\frac{\partial M}{\partial y}\) equals \(2 \frac{\partial N}{\partial x}\), the equation is not exact, but homogeneous, which can be solved by substitution method using \(v = \frac{y}{x}\)

Use substitution method (v = y/x)

We are given the equation: \[(x + 4) (y^2 +1) dx + 2y (x^2 + 3x + 2) dy = 0\] Let's substitute \(v = \frac{y}{x}\), which implies \(y = vx\) and \(\frac{dy}{dx} = v + x \frac{dv}{dx}\): \[ (x + 4) (v^2x^2+1) + 2vx(x^2+3x+2) \left( v + x\frac{dv}{dx}\right) = 0 \] Now, we need to solve the resulting equation for \(\frac{dv}{dx}\):

Solve for dv/dx

From the previous step, we have: \[ (x + 4)(v^2x^2 + 1) + 2vx(x^2 + 3x + 2) \left( v + x\frac{dv}{dx}\right) = 0 \] Rearrange the equation to isolate \(\frac{dv}{dx}\): \[ x\frac{dv}{dx} = - \frac{(x + 4)(v^2x^2 + 1) + 2v^2x(x^2 + 3x + 2)}{2v(x^2 + 3x + 2)} \] Now, integrate both sides with respect to x: \[\int \frac{dv}{dx} dx = - \int \frac{(x + 4)(v^2x^2 + 1) + 2v^2x(x^2 + 3x + 2)}{2v(x^2 + 3x + 2)} dx\]

Find the general solution

This step involves the integration of both sides of the equation: \[\int \frac{dv}{dx} dx = - \int \frac{(x + 4)(v^2x^2 + 1) + 2v^2x(x^2 + 3x + 2)}{2v(x^2 + 3x + 2)} dx\] \[\int dv = - \int \frac{(x + 4)(v^2x^2 + 1) + 2v^2x(x^2 + 3x + 2)}{2v(x^2 + 3x + 2)} dx\] \[v = - \int \frac{(x + 4)(v^2x^2 + 1) + 2v^2x(x^2 + 3x + 2)}{2v(x^2 + 3x + 2)} dx + C\] Since \(v = \frac{y}{x}\), we have: \[\frac{y}{x} = - \int \frac{(x + 4)(v^2x^2 + 1) + 2v^2x(x^2 + 3x + 2)}{2v(x^2 + 3x + 2)} dx + C\] This is the general solution of the given differential equation.

Key concepts

Homogeneous Differential Equations

Homogeneous differential equations are a special kind of equations that can often be solved efficiently by a method called substitution. These equations are characterized by being expressed such that every term of the equation has the same degree when substituting variables. This property allows you to transform the equation into a simpler form, facilitating easier solutions. In practice, a differential equation is considered homogeneous if it can be expressed in terms of a function where every component is proportional to some power of the state variables.

For instance, if you have an equation where each term can be factored uniformly, or scales in the same way, when applying independent scaling terms to both the dependent and independent variables, it's likely homogeneous. However, not every differential equation is naturally homogeneous. Often, you'll find these equations in forms that need manipulation such as factoring or leveraging exact differentials to bring them to a homogeneous standard.

Homogeneous differential equations frequently appear in dynamic systems and various scientific models. It's often crucial for problem solvers to recognize this form because it leads to elegant approaches in derivation and analysis, markedly shortening and simplifying the process of finding a solution.

Substitution Method in Differential Equations

Substitution methods are powerful tools in the world of differential equations, particularly valuable in dealing with complex relationships between variables. With homogeneous differential equations, substitution essentially allows the transformation of the equation into a solvable form. In our example problem, the substitution begins by setting \(v = \frac{y}{x}\), which implies the relationship between \(y\) and \(x\) is dependent on the ratio rather than their specific values.

By doing this substitution, complicated differential equations that involve two variables become simpler, often transforming into a linear differential equation. The main goal is to simplify the terms to a point where traditional analytical techniques, like integration, can be effectively applied.

• Identify the substitution that can simplify the equation.
• Replace the original variables with the new substitute variables.
• Transform the derivative terms accordingly using the chain rule.
• Rearrange the transformed equation to find a relative differential equation.

The result is usually a function in terms of the new substituted variable, which can then be reintegrated back into the original variable context. This substitution enables a clear pathway to solution, which previously may have been obscured by complex interdependencies in the original form.

Integration Techniques in Differential Equations

Integration is a fundamental step in solving differential equations, once we've employed the necessary substitutions to simplify the equation. The integration process essentially involves finding an antiderivative, or the function whose derivative yields the terms we have.

In the exercise problem, after substituting and rearranging the terms, we transform the differential equation into a form amenable to integration. Here's where a variety of integration techniques can come into play, such as:

• Integration by parts, especially when dealing with products of functions.
• Substitution, again, if a more suitable variable change emerges within the integral.
• Partial fraction decomposition, if the integrand displays a rational function.

The integration constants that arise are crucial as they reflect initial conditions or constraints in physical and geometric contexts. In our example, integrating both sides and solving for the substitution gives the solution in terms of \(v\), which ultimately needs to be expressed back in terms of \(x\) and \(y\) through \(v = \frac{y}{x}\).

Thus, the integration techniques bridge the gap from having a complex equation to achieving a usable, concrete solution, which is essential for practical applications and understanding the underlying behavior described by the differential equation.