Question

Find all solutions. $$ y^{\prime} \ln |y|+x^{2} y=0 $$

Short answer

Answer: The general solution of the given differential equation is $$y(x) = \pm e^{e^{-\frac{1}{3}x^3 + C}}$$.

Step-by-step solution

Rewrite the differential equation

We rewrite the given differential equation as: $$ \frac{dy}{dx} \ln |y| + x^2 y = 0 $$

Transform the equation into an exact equation

We can treat $$\frac{dy}{dx}$$ as a fraction and rearrange the equation: $$ \frac{dy}{y\ln|y|} = -x^2 dx $$ Now, we can integrate both sides of the equation.

Integrate both sides

Integrating both sides with respect to their respective variables, we get: $$ \int \frac{1}{y\ln|y|}dy = -\int x^2 dx + C $$ On the left side, we can use substitution method, let $$u = \ln|y|$$, then $$du = \frac{1}{y}dy$$. On the right side, integrate $$-x^2$$ with respect to x. So, the equation becomes: $$ \int \frac{1}{u} du = -\int x^2 dx + C $$ Integrating each side, we obtain: $$ \ln|u| = -\frac{1}{3}x^3 + C $$ Now, substitute back $$u = \ln|y|$$: $$ \ln|\ln|y|| = -\frac{1}{3}x^3 + C $$

Determine the general solution

To find the general solution, we need to get y by itself. First, we can rewrite as: $$ \ln|y| = e^{-\frac{1}{3}x^3 + C} $$ Applying exponent rule, we get: $$ y = \pm e^{e^{-\frac{1}{3}x^3 + C}} $$ In the given notation, the general solution of the given differential equation is: $$ y(x) = \pm e^{e^{-\frac{1}{3}x^3 + C}} $$

Key concepts

Differential Equation Integration

Differential equation integration is at the heart of solving many types of differential equations. It is the process of finding a function whose derivative matches the given differential equation. The purpose of integration here is to 'undo' differentiation, which in theory, allows us to retrieve the original function from its derivative.

For the given exercise, the integration process begins after transforming the differential equation into a separable form, which leads to the integration of two distinct parts. The integration process is the crucial step towards finding the general solution, where integral calculus is used to systematically determine the antiderivatives of the functions on both sides of the equation.

Separable Differential Equations

Separable differential equations are a class of differential equations that can be simplified into two separate functions, one involving only the variable x and the other involving only the variable y. The original equation is therefore 'separated' into two parts that can be integrated independently of each other.

In the problem at hand, the equation is transformed by dividing through by terms involving y, allowing the variables to be separated. When you see an equation like \(\frac{dy}{dx} = g(x)h(y)\), you can rearrange it to \(\frac{dy}{h(y)} = g(x)dx\), which allows you to integrate both sides with respect to their respective variables. This step effectively divides the problem into smaller, more manageable parts.

Exact Differential Equations

An exact differential equation is a specific type of differential equation where the expressions on both sides of the equation can be identified as the derivatives of some function with respect to x and y. However, not all differential equations are exact, and the one in our exercise is not naturally exact but can be made separable.

The concept of exactness is more relevant in cases where the initial equation is a perfect differential, or where clever manipulation using integrating factors turns a non-exact equation into an exact one. While important, exact differential equations do not directly apply to our exercise since we are not dealing with an explicitly exact differential equation nor using an integrating factor.

Integration by Substitution

Integration by substitution is a technique often referred to as u-substitution. It's used when an integral contains a function and its derivative, allowing a complex expression to appear simpler and thus easier to integrate.

In the context of our problem, integration by substitution comes into play when we're tasked with integrating \(\int \frac{1}{y\ln|y|}dy\). By setting \(u = \ln|y|\) and \(du = \frac{1}{y}dy\), we transform the integral into a more manageable form, \(\int \frac{1}{u} du\), which is a straightforward integral leading us directly to \(\ln|u|\). Through substitution, we've circumvented the complexity of the original integral and streamlined the path to a solution.