Question

Solve the initial value problem $$ y^{\prime}=3 x^{2} /\left(3 y^{2}-4\right), \quad y(1)=0 $$ and determine the interval in which the solution is valid. Hint: To find the interval of definition, look for points where the integral curve has a vertical tangent.

Short answer

To summarize, the solution of the given initial value problem is \(y(x) = \sqrt{\frac{4(e^{6x^3-1})}{3}}\), and the interval of validity for this solution is \(x \in (x_1, x_2)\), where \(x_1 = \sqrt[3]{\frac{1}{6} \ln\left(\frac{12}{3}\right)}\) and \(x_2 = \sqrt[3]{\frac{1}{6} \ln\left(\frac{12}{-3}\right)}\).

Step-by-step solution

Separate Variables

Divide both sides by \(3y^2-4\) and multiply both sides by \(dy\) to obtain $$ \frac{dy}{3y^2-4} = 3x^2 dx. $$

Integrate Both Sides

Integrate both sides with respect to their respective variables to obtain $$ \int \frac{dy}{3y^2-4} = \int 3x^2 dx. $$ For the left-hand side, perform a substitution \(u=3y^2-4 \Rightarrow du=6y\;dy\), giving $$ \frac{1}{6}\int \frac{du}{u} = \int 3x^2 dx. $$ Now we can evaluate both integrals: $$ \frac{1}{6} \ln|u| = x^3 + C_1. $$ Substitute back \(u=3y^2-4\) to get $$ \frac{1}{6} \ln|3y^2-4| = x^3 + C_1. $$

Use Initial Condition and Solve for \(y\)

Apply the initial condition, \(y(1)=0\), to find the value of \(C_1\): $$ \frac{1}{6} \ln|3(0)^2-4| = 1^3 + C_1, $$ which gives \(C_1=-\frac{1}{6}\ln 4\). Now we can rewrite the equation as $$ \frac{1}{6} \ln|3y^2-4| = x^3 - \frac{1}{6} \ln 4. $$ Solve for \(y(x)\): $$ y(x) = \sqrt{\frac{4(e^{6x^3-1})}{3}}. $$

Determine the Interval of Validity

Recall the hint: to find the interval of definition, look for points where the integral curve has a vertical tangent. In other words, the interval will be valid from the point where \(y'(x) = \infty\). The given differential equation is $$ y^{\prime}=3 x^{2} /\left(3 y^{2}-4\right). $$ Setting \(3y^2-4\) equal to zero and solving for \(y\) gives us $$ y = \pm\frac{2}{\sqrt{3}}. $$ Now, we need to find the \(x\)-values corresponding to these \(y\) values: $$ x_1 = \sqrt[3]{\frac{1}{6} \ln\left(\frac{12}{3}\right)},\quad x_2 = \sqrt[3]{\frac{1}{6} \ln\left(\frac{12}{-3}\right)}. $$ Thus, the interval in which the solution is valid is \(\boxed{x \in (x_1, x_2)}\).

Key concepts

Separation of Variables

When faced with a differential equation, one technique often used for finding its solution is the separation of variables. This method involves rearranging the terms of the equation so that each variable and its differential are on opposite sides of the equation. The basic idea is to treat the differentials \(dx\), \(dy\) as separate entities that can be manipulated algebraically.

For the given initial value problem \(y' = 3x^{2}/(3y^{2}-4)\), which is a first-order ordinary differential equation, we begin the separation process by dividing both sides by \(3y^2-4\) and multiplying both sides by \(dy\). This results in isolated differentials that set the stage for integration, which is our next critical step in solving the equation:
\[\frac{dy}{3y^2-4} = 3x^2 dx.\]
By separating variables, we've effectively restructured the equation in such a way that we can now integrate each side with respect to its own variable.

Integrating Differential Equations

After separating the variables of a differential equation, we proceed to the integration phase. Integration is a fundamental tool in calculus, central to solving differential equations. We integrate to find the antiderivatives of the separated functions which, in turn, provides us with the general solution to the differential equation.

In our example, separate integrations for \(y\) and \(x\) are performed on the separated equation. Substitution is used on the lefthand side to manage the integration:\

• Let \(u = 3y^2 - 4\) and \(du = 6y dy\), so \(dy = \frac{du}{6y}\).
• The integral transforms into \(\frac{1}{6}\int \frac{du}{u}\), which simplifies to \(\frac{1}{6}\ln|u|\).

Following integration, we include the constant of integration, which is the 'C' seen in definite integrals. It is essential for representing the family of all possible solutions. The result is combined into a single equation reflecting the relationship between \(x\) and \(y\), leading us toward the explicit solution after applying initial conditions.

Interval of Validity

The interval of validity for the solution to a differential equation represents the range of the independent variable—usually \(x\), in our case—where the solution is defined and complies with the initial conditions. For the differential equation at hand, it's important to find where the solution might not be valid, such as where division by zero might occur or the function becomes discontinuous.

In the sample problem, the integral curve of the differential equation may have a vertical tangent where the derivative \(y'\) is infinite. This typically occurs when the denominator in the original differential equation is zero. Setting \(3y^2-4=0\) and solving for \(y\), we identify potential points of discontinuity for \(y\).
\[y = \pm\frac{2}{\sqrt{3}}.\]
Next, we find corresponding \(x\)-values to determine the actual interval of validity. The solution remains valid between these \(x\)-values, which gets identified by considering points close to the vertical tangents. Recognizing the interval of validity ensures that we understand the limitations and appropriateness of our solution within its specific context.