Question

In Exercises \(61-68\) use the method suggested by Exercise 60 to find the general solution on some interval \((0, \rho)\) $$ 6 x^{2}\left(1+2 x^{2}\right) y^{\prime \prime}+x\left(1+50 x^{2}\right) y^{\prime}+\left(1+30 x^{2}\right) y=0 $$

Short answer

Question: Find the general solution of the given second-order differential equation on an interval \((0, \rho)\): $$6x^2(1+2x^2)y'' + xy(1+50x^2)y' + (1+30x^2)y = 0$$. Solution: By solving the given Cauchy-Euler equation, we find that the general solution is: $$y(x) = C_1x^{\frac{1}{3}} + C_2x^2$$, where \(C_1\) and \(C_2\) are constants.

Step-by-step solution

Assume the solution form and compute the derivatives

Assume that the solution has the form of \(y=x^r\) for some constant r. Calculate the first and second derivatives of y with respect to x: $$y'=rx^{r-1}$$ $$y''=r(r-1)x^{r-2}$$

Substitute the solution and derivatives into the given equation

Substitute y, y', and y'' into the given differential equation: $$6x^2(1+2x^2)r(r-1)x^{r-2} + x(1+50x^2)rx^{r-1} + (1+30x^2)x^r = 0$$

Simplify the equation

Factor out \(x^r\) from the equation and simplify: $$x^r\left[6r(r-1)(1+2x^2) + r(1+50x^2) + (1+30x^2)\right]=0$$ Since \(x^r\neq 0\), we can divide both sides of the equation by \(x^r\), giving: $$6r(r-1)(1+2x^2) + r(1+50x^2) + (1+30x^2) = 0$$

Solve for r

Set the terms in the brackets to zero and solve for r: $$6r^2 - 6r + r + 1 = 0$$ $$6r^2 - 5r + 1 = 0$$ Now solve the quadratic equation for r: $$r = \frac{5 \pm \sqrt{5^2 - 4(6)(1)}}{2(6)}$$ $$r = \frac{1}{3}, 2$$

Write the general solution

Now that we have the two values for r, we can write down the general solution to the given equation as a linear combination of \(y_1\) and \(y_2\): $$y(x) = C_1x^{\frac{1}{3}} + C_2x^2$$ Therefore, the general solution of the given differential equation on an interval \((0, \rho)\) is: $$y(x) = C_1x^{\frac{1}{3}} + C_2x^2$$

Key concepts

General Solution

When we talk about the "general solution" of a differential equation, we are referring to a formula that contains all possible solutions of the equation. These solutions are usually expressed in terms of arbitrary constants. In this context, the arbitrary constants are represented by symbols like \(C_1\) and \(C_2\). These constants can take on any value, making the solution "general" since it covers a wide range of specific solutions.

In our exercise, the general solution is given as a linear combination of two separate solutions: \(y_1 = x^{\frac{1}{3}}\) and \(y_2 = x^2\). The linear combination is written as:

• \(y(x) = C_1 x^{\frac{1}{3}} + C_2 x^2\)

This form allows us to encompass a multitude of specific solutions based on the chosen values for \(C_1\) and \(C_2\).

Understanding the general solution is crucial because it gives a comprehensive insight into the behavior of differential equations on different intervals.

Quadratic Equation

Quadratic equations are polynomial equations of the form \(ax^2 + bx + c = 0\), where \(a\), \(b\), and \(c\) are constants. Solving these equations involves finding the values of \(x\) that make the equation true. Often, the quadratic formula is used:

• \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\)

In our exercise, a quadratic equation emerges when we seek to find the values of \(r\) that satisfy the differential equation. By substituting into the original equation, we derive the quadratic equation:

• \(6r^2 - 5r + 1 = 0\)

Solving this using the quadratic formula, we identified two solutions: \(r = \frac{1}{3}\) and \(r = 2\).

These roots are critical to forming the general solution, as each root corresponds to a power of \(x\) in the solution.

Interval Notation

Interval notation is a mathematical shorthand used to describe sets of numbers or intervals. It makes it easier to denote ranges of values concisely. For example, the interval notation \((0, \rho)\) indicates all numbers between 0 and \(\rho\), excluding 0 and including \(\rho\).

In our exercise, the general solution is expressed over the interval \((0, \rho)\), which is necessary due to the nature of the solutions. Sometimes, differential equations have solutions that are only valid over specific domains where certain conditions, such as continuity or differentiability, are met.

Understanding interval notation helps in correctly applying solutions to specific parts of the number line, ensuring that the solution remains valid.

Linear Combination

The concept of a "linear combination" involves merging two or more functions, scaling each by a constant. In our exercise, the general solution is expressed as a linear combination of two independent solutions \(y_1(x) = x^{\frac{1}{3}}\) and \(y_2(x) = x^2\). By pulling these solutions together into a single formula, we construct the general solution:

• \(y(x) = C_1 x^{\frac{1}{3}} + C_2 x^2\)

Here, \(C_1\) and \(C_2\) are the constants that allow the solutions to be adjusted based on initial conditions or specific requirements.

Linear combinations are widely used to express general solutions in differential equations, allowing for combinations of functions to cover the complete spectrum of possible solutions.