Question

Solve the equation $$ \left(x^{2}+y^{2}\right) y^{\prime}+x y=0 $$ by using the change of variable \(v=y / x\) and \(v=x / y\).

Short answer

Question: Determine the general solution to the given first-order ordinary differential equation (ODE) using the change of variables \(v = y/x\) and \(v = x/y\): $$ (x^2 + y^2)y' + xy = 0. $$ Answer: Unfortunately, the integral on the left-hand side of the resulting equation cannot be expressed in elementary functions. As a result, we can only express the general solution in terms of an integral function \(F(v)\): $$ F(v) = -\int \dfrac{1}{x(x^2 + 1)} dx + C, $$ where \(C\) is the constant of integration. To find the particular solution, additional techniques could be required, and you would need to be given an initial condition.

Step-by-step solution

Apply the change of variable \(v = y/x\) and find the derivative

Let's begin by applying the change of variable \(v = y/x\). We have \(y = vx\) and we need to find its derivative with respect to \(x\), which we'll denote as \(y'\). Using the chain rule, we have: $$ y^{\prime} = v^{\prime} x + v $$ Now we have a new equation for \(y'\) in terms of \(v\) and \(x\), which we will substitute back into the original ODE.

Substitute the new equation into the original ODE

Now, we substitute the new equation for \(y\) and \(y'\) back into the original ODE: $$ \left(x^2 + (vx)^2 \right)(v^{\prime}x + v) + x(vx) = 0. $$ Let's simplify the equation: $$ (x^2 + v^2x^2)(v^{\prime}x + v) + xv^2x = 0. $$ We can divide both sides of the equation by \(x\) to get rid of the common factor: $$ (x + v^2x)(v^{\prime}x + v) + v^2x = 0. $$

Solve the new ODE

The equation now simplifies to: $$ (x(v^2+1))(xv'+v)+v^2x=0 $$ Now let's group the terms with \(x\) together: $$ x^2(v^2+1)v'+x(v^3+2v)+v^2x=0 $$ Now, we can simplify the equation: $$ x^2(v^2+1)v'+x(v^3+2v+v^2)=0. $$ Next, we separate the variables in the equation: $$ \dfrac{v'}{v^3+v^2+2v} = -\dfrac{1}{x(x^2+1)}. $$ Now, we will integrate both sides of the equation:

Integrate both sides and find the general solution

Integrate both sides of the equation: $$ \int \dfrac{1}{v^3 + v^2 + 2v} dv = -\int \dfrac{1}{x(x^2 + 1)} dx. $$ Unfortunately, the integral on the left-hand side cannot be expressed in elementary functions. However, you can denote the indefinite integral as a new function, say \(F(v)\): $$ F(v) = -\int \dfrac{1}{x(x^2 + 1)} dx + C, $$ where \(C\) is the constant of integration. To find the particular solution, you would need to be given an initial condition, and additional techniques could be required to evaluate these integrals. From this point, we could apply the second change of variable to arrive at a solution.

Key concepts

Change of Variable

The strategy of change of variable is a powerful mathematical tool used to simplify complex differential equations. It works by replacing one variable with another, which often transforms a difficult equation into a more manageable form. In the context of ordinary differential equations (ODEs), change of variable can make an intractable equation solvable.

For instance, given an equation involving variables x and y, we might introduce a new variable v that is defined as a function of x and y, such as \(v = y/x\). This could simplify the equation by reducing the number of terms or by turning a non-linear equation into a linear one. Transforming an ODE using a change of variable is often the first step to finding a solution.

Chain Rule

The chain rule is a fundamental principle in calculus that is used when differentiating composite functions. It states that the derivative of a composite function is the derivative of the outer function evaluated at the inner function, multiplied by the derivative of the inner function.

In the context of solving ODEs, the chain rule allows us to differentiate a function that has been defined in terms of another variable. If we have \(y = v(x) * x\), we apply the chain rule to obtain \(y' = v' * x + v\) when finding \(y'\), the derivative of y with respect to x. This technique is critical when employing a change of variable to an ODE, because it allows us to express derivatives in terms of the new variable.

Separation of Variables

The method of separation of variables is a technique used to solve separable ODEs. This method involves rearranging the terms of a differential equation so that each variable and its derivatives are on opposite sides of the equation.

To implement this technique, we assume the ODE can be written in the form \(g(y) * dy = f(x) * dx\), allowing us to integrate both sides separately with respect to their own variables. The separated integrals lead to the general solution of the ODE, often involving an integration constant. The method of separation of variables is an efficient way to solve many first-order ODEs and is especially relevant when we have performed a suitable change of variable beforehand.

Integration of ODE

After applying the aforementioned techniques like change of variable and separation of variables, the next step is the integration of ODE. Integration is the inverse process of differentiation, and it is employed to find the antiderivative or integral of an equation.

When solving a differential equation, we integrate both sides of the equation after separation of variables to find the original function y from its derivative \(y'\). The integration process usually introduces an arbitrary constant, known as the constant of integration, denoted by \(C\). This constant can later be determined if an initial condition is provided, leading to a particular solution of the ODE.

Variable Substitution in ODEs

In the process of solving differential equations, variable substitution is often utilized in conjunction with the integral calculus to simplify and solve an ODE. Such substitution not only can involve the change of dependent variables but also the independent variables.

Effective substitution transforms the original ODE into an equation that is easier to integrate. For example, when faced with a challenging integral as part of an ODE solution, an appropriate substitution might convert it into a standard form, for which the antiderivative is known. While the substitution method is highly versatile, it requires careful selection of the new variables and a good understanding of their relationships to the original variables.