Question

$$ y^{\prime \prime}-3 y^{2}=0, \quad y(0)=2, \quad y^{\prime}(0)=4 $$

Short answer

Question: Rewrite the given second order non-linear ordinary differential equation as a system of two first order ODEs and find the implicit expression for the solution using the given initial conditions: $$ y^{\prime \prime} - 3 y^2 = 0, \quad y(0) = 2, \quad y^{\prime}(0) = 4 $$ Answer: The implicit expressions for the solution are: $$ y_1(t) = \int y_2(t) dt + 2 \quad \text{and} \quad y_2(t) = \int 3 y_1^2(t) dt - 8, $$ where \(y(t) = y_1(t)\) and \(y^{\prime}(t) = y_2(t)\).

Step-by-step solution

Rewrite the second order ODE as two first order ODEs

Let \(y_1 = y(t)\) and \(y_2 = y^{\prime}(t) = y_1^{\prime}(t)\). So the given equation becomes: $$ y_2^{\prime} - 3 y_1^2 = 0 $$ Now we have a set of two first order ODEs: $$ \begin{cases} y_1^{\prime} = y_2 \\ y_2^{\prime} = 3 y_1^2 \end{cases} $$

Solve the first order ODE for \(y_1\)

The integral of the first equation is straightforward: $$ y_1(t) = \int y_2(t) dt + C $$ Use the initial condition \(y_1(0) = 2\): $$ 2 = \int y_2(0) dt + C \implies C = 2 $$ So we get: $$ y_1(t) = \int y_2(t) dt + 2 $$

Solve the first order ODE for \(y_2\)

The integral of the second equation is: $$ y_2(t) = \int 3 y_1^2(t) dt + C $$ However, this integral is a bit difficult to solve analytically for the general solution, so we need to use the initial condition \(y_2(0) = 4\) to get an equation for the specific solution: $$ 4 = \int 3 (y_1(0))^2 dt + C \implies C = 4 - 3\times(2)^2 = -8 $$ So, the specific solution for \(y_2(t)\) is given by: $$ y_2(t) = \int 3 y_1^2(t) dt - 8 $$ These equations for \(y_1(t)\) and \(y_2(t)\), together with the relation \(y(t)=y_1(t)\) and \(y^{\prime}(t)=y_2(t)\), give an implicit expression for the solution to the problem. However, finding a more explicit solution would require a more complex approach, potentially involving numerical methods or approximations.

Key concepts

Initial Value Problem

An initial value problem (IVP) in the context of differential equations refers to finding a function that satisfies a given ordinary differential equation (ODE) and also meets specified initial conditions. The classic form of an IVP is \( y'(t) = f(t, y(t)) \), with initial conditions given that typically establish the value of \( y(t) \) at a certain point, like \( y(t_0) = y_0 \).

In our exercise, we're dealing with a second-order ODE, which can be rewritten as a system of first-order ODEs. To comply with the IVP requirements, we have initial conditions \(y(0) = 2\) and \(y'(0) = 4\). These conditions allow us to determine the constants of integration when solving the system of ODEs step by step.

First Order ODE Systems

When solving higher-order ODEs, like second-order differential equations, a common strategy is to convert them into systems of first-order ODEs. This step is crucial because it breaks down a complex equation into a system that is more manageable and compatible with various solution methods.

In our original exercise, the second-order ODE \( y'' - 3y^2 = 0 \) is transformed into a system of first-order ODEs. We let \( y_1 = y \) and \( y_2 = y' \), and then we can express the original second-order ODE as \( y_1' = y_2 \) and \( y_2' = 3 y_1^2 \). This form is much more tractable and paves the way for using techniques oriented to systems of equations, like matrix methods or phase plane analysis.

Integral Solutions

Integral solutions involve finding the antiderivatives of functions, which is a fundamental aspect of solving differential equations. The integration process takes a derivative and recovers the original function, or at least a family of potential original functions.

In solving for \( y_1(t) \) and \( y_2(t) \) in our exercise, we encounter integrals that need to be evaluated. The initial conditions become pivotal because they allow us to solve for the constants that appear as a result of indefinite integration. For example, the integral solution for \( y_1 \) uses the condition \( y_1(0) = 2 \) to find that one integral solution is \( y_1(t) = \int y_2(t) dt + 2 \) while for \( y_2 \) we incorporate \( y_2(0) = 4 \) to refine the family of solutions down to a specific one.

Numerical Methods

There are scenarios where an ODE or a system of ODEs cannot be solved analytically or where finding an explicit formula for the solution is exceedingly difficult. In such cases, numerical methods come to the rescue. These methods provide approximate solutions to differential equations within a desired degree of accuracy.

Common numerical methods include Euler's method, Runge-Kutta methods, and predictor-corrector methods. The choice among these depends on factors such as the desired precision and the computational resources available. In the exercise mentioned earlier, after the initial attempt to find a closed-form solution, one may turn to numerical methods for an approximate solution that still captures the necessary dynamics of the original problem.