Question

Find the solution to the initial-value problem.\(y^{\prime}=3 x^{2}\left(y^{2}+4\right), y(0)=0\)

Short answer

The solution is \( y = 2\tan(2x^3) \).

Step-by-step solution

Identify the Type of Differential Equation

The given differential equation is a first-order, non-linear ordinary differential equation: \( y' = 3x^2(y^2 + 4) \). Moreover, it is a separable differential equation because the expression can be rewritten such that all terms involving \( y \) are on one side of the equation and all terms involving \( x \) are on the other side.

Separate the Variables

Rewrite the equation to separate variables:\[ \frac{dy}{y^2 + 4} = 3x^2 \, dx \].This allows us to integrate each side with respect to their respective variables.

Integrate Both Sides

Integrate both sides of the equation. The left side can be integrated to\[ \int \frac{dy}{y^2 + 4} = \frac{1}{2} \arctan\left(\frac{y}{2}\right) + C_1 \].The right side is:\[ \int 3x^2 \, dx = x^3 + C_2 \].Thus, after integration, we have:\[ \frac{1}{2} \arctan\left(\frac{y}{2}\right) = x^3 + C. \]

Solve for the Integration Constant

Use the initial condition \( y(0) = 0 \) to find the constant \( C \). Substituting \( x = 0 \) and \( y = 0 \) into the integrated equation gives:\[ \frac{1}{2} \arctan(0) = 0^3 + C \].Since \( \arctan(0) = 0 \), it follows that \( C = 0 \).

Solve for y in terms of x

Now, we solve \[ \frac{1}{2} \arctan\left(\frac{y}{2}\right) = x^3 \] for \( y \):Multiply both sides by 2:\[ \arctan\left(\frac{y}{2}\right) = 2x^3 \].Take the tangent of both sides:\[ \frac{y}{2} = \tan(2x^3) \].Finally, solve for \( y \):\[ y = 2\tan(2x^3) \].

Key concepts

Initial Value Problem

An initial value problem in differential equations provides a plus to overcome ambiguity. It involves not just the differential equation itself, but also an initial condition. This condition, usually of the form \( y(x_0) = y_0 \), serves to pinpoint the unique solution to the differential equation that respects this starting point. Here the problem is:

• **Differential Equation:** \( y' = 3x^2(y^2 + 4) \)
• **Initial Condition:** \( y(0) = 0 \)

The initial condition anchors the solution at a specific point, ensuring that out of the many possible functions that could satisfy the differential equation, we choose the one that intersects this point. This is particularly vital when dealing with non-linear equations, where solutions aren't straightforward.

Separable Differential Equations

Separation of variables is a method for solving differential equations. It is applicable when the variables can be separated on each side of the equation. For the given equation, \( y' = 3x^2(y^2 + 4) \), this means we can rearrange terms to isolate \( y \) and \( x \):
Firstly, you rearrange to get:
\[ \frac{dy}{y^2 + 4} = 3x^2 \, dx \] This separates the terms without intermixing them. The left side now features only \( y \), and the right side only \( x \).
Once the variables are separated, each side can be independently integrated, allowing the solution to be determined step by step.

Non-linear Differential Equations

Non-linear differential equations are equations where the dependent variable or its derivatives are raised to a power greater than one or appear as products with other terms. The given equation, \( y' = 3x^2(y^2 + 4) \), is non-linear due to the \( y^2 \) term.
Such equations often don't have solutions that can be expressed as simple combinations of basic functions like polynomials or trigonometric functions. Instead, they might yield more complex expressions or even require numerical approximations for solutions.
Working with non-linear equations can involve various mathematical techniques, with separation of variables being one handy method for certain forms of these equations.

Integration of Functions

When solving a differential equation by separation of variables, each side of the equation is integrated separately. Integration transforms a derivative back into its original function. For the components:- **Left Side Integration:** We have \( \int \frac{dy}{y^2 + 4} \), which uses a known integral formula, resulting in \( \frac{1}{2} \arctan\left(\frac{y}{2}\right) + C_1 \).- **Right Side Integration:** Integration of \( \int 3x^2 \, dx \) yields \( x^3 + C_2 \).After integrating both sides, constants \( C_1 \) and \( C_2 \) can be amalgamated as a single constant \( C \), aiding the simplification and ensuring continuity over the entire function.
Integrating functions helps reverse the differentiation process, bringing us closer to finding an explicit solution for the differential equation.