Question

Solve the given differential equation. $$ y^{\prime}=x^{2} / y\left(1+x^{3}\right) $$

Short answer

Short Answer: To solve the given first-order differential equation \(y^{\prime} = \frac{x^2}{y(1+x^3)}\), we first separate the variables, giving us \(y y^{\prime} = \frac{x^2}{1+x^3}\). After integrating both sides with respect to x, we obtain \(\frac{1}{2}y^2 = \frac{1}{3}\ln|1+x^3| + C\). Solving for y, we find the general solution: \(y(x) = \pm\sqrt{2\left(\frac{1}{3}\ln|1 + x^3| + C\right)}\).

Step-by-step solution

Separate the variables

We need to rewrite the given equation to separate the variables x and y. Multiply both sides of the equation by y(1 + x^3) to get: $$ y(1+x^{3})y^{\prime} = x^{2} $$ Now divide both sides by (1 + x^3): $$ y y^{\prime} = \frac{x^{2}}{1 + x^{3}} $$

Integrate both sides of the equation

Now, integrate both sides of the equation with respect to x: $$ \int y y^{\prime}\, dx = \int \frac{x^{2}}{1 + x^{3}}\, dx $$ Notice that the left side of the equation is the derivative of \(\frac{1}{2}y^2\). So, we can write the equation as: $$ \frac{1}{2}y^2 = \int \frac{x^{2}}{1 + x^{3}}\, dx + C $$

Solve for y

To find the explicit solution, we will first solve the integral on the right side: $$ \int \frac{x^{2}}{1 + x^{3}}\, dx $$ Substitute u = 1 + x^3, then du = 3x^2 dx: $$ \frac{1}{3}\int \frac{du}{u} $$ Now we can integrate: $$ \frac{1}{3}\ln|u| = \frac{1}{3}\ln|1 + x^{3}| $$ Substitute back the value of u and rearrange the equation: $$ \frac{1}{2}y^2 = \frac{1}{3}\ln|1 + x^{3}| + C $$ To solve for y, take the square root of both sides: $$ y(x) = \pm\sqrt{2\left(\frac{1}{3}\ln|1 + x^{3}| + C\right)} $$ This is the general solution of the given differential equation.

Key concepts

Separation of Variables

One fundamental method to tackle differential equations is the separation of variables. This approach is highly useful whenever a differential equation can be rewritten such that each variable (and its derivative) is on opposite sides of the equation. For example, if you start with an equation like \( y^{\prime}=x^{2} / y(1+x^{3}) \), the goal is to manipulate the equation until you have all the terms involving \( y \) on one side and terms involving \( x \) on the other. In the case of our equation, multiplications and divisions are usually used to achieve this separation. By doing so, the equation becomes more manageable and ready for the next step, which is integration. You'll end with something like \( y y^{\prime} = \frac{x^{2}}{1 + x^{3}} \), ensuring a distinct division between variables.

Integration Techniques

Integration is a critical skill when working with differential equations and requires using various techniques to solve the integrals effectively. Once the differential equation is separated into two integrals according to their respective variables, each side is integrated. For the equation \( y y^{\prime} = \frac{x^{2}}{1 + x^{3}} \), integrating the left side, \( y y^{\prime} \), results in \( \int y y^{\prime}\, dx = \frac{1}{2}y^2 \). On the right side, the integral \( \int \frac{x^{2}}{1 + x^{3}}\, dx \) may require a substitution technique. In this instance, substituting \( u = 1 + x^3 \) simplifies the integral to a basic logarithmic form \( \frac{1}{3}\ln|u| \). Recognizing when to use certain techniques, such as substitution or recognizing standard forms, is essential for solving complex integrals.

General Solution

The ultimate goal when solving a differential equation is to find a general solution, which includes all possible solutions of the equation. After applying separation of variables and solving the integrals, you'll reach an expression like \( \frac{1}{2}y^2 = \frac{1}{3}\ln|1 + x^{3}| + C \). Here, \( C \) is an arbitrary constant representing the general solution's family of curves. Solving this expression for \( y \) gives us \( y(x) = \pm\sqrt{2\left(\frac{1}{3}\ln|1 + x^{3}| + C\right)} \). The square root indicates two potential solutions due to the \( \pm \) sign, showcasing the richness and variability of differential equations' solutions. Hence, the general solution doesn't just provide a single pathway but rather envelops a range of possibilities governed by initial conditions.