Determine three linearly independent solutions to the given...

Determine three linearly independent solutions to the given differential equation of the form $y(x)=x^{r},$ and thereby determine the general solution to the differential equation on $(0, \infty)$.
$$x^{3} y^{\prime \prime \prime}+3 x^{2} y^{\prime \prime}-6 x y^{\prime}=0, x > 0$$

Key Concepts

Indicial Equation

The indicial equation is an algebraic equation obtained by substituting y(x) = x^r into the differential equation. The roots of this equation determine the exponents in the power-law solutions. These roots, when distinct and ensuring linear independence, provide the basis for constructing the general solution to the differential equation.

General Solution of Linear Differential Equations

The general solution of a linear differential equation is formed by a linear combination of linearly independent solutions. Once a fundamental set of solutions is found, such as those obtained using the power-law substitution method and solving the indicial equation, any solution of the differential equation can be expressed as a weighted sum of these basis functions. This approach ensures that the solution covers the entire solution space appropriate for the differential equation on the given interval.

Euler-Cauchy Differential Equation

An Euler-Cauchy differential equation is a type of linear differential equation in which each derivative is multiplied by a power of the independent variable. This specific structure allows the use of a substitution like y = x^r to transform the variable coefficient equation into an algebraic equation, thereby simplifying the process of finding its solutions.

Power-Law Substitution Method

The power-law substitution method involves assuming a solution of the form y(x) = x^r. This method is particularly effective for Euler-Cauchy equations because it converts the differential equation into an algebraic (or indicial) equation in the exponent r. The resulting equation can then be solved for the values of r which correspond to potential solutions of the original differential equation.

Transcript

00:01 So for this differential equation, we need to find solutions of the form y is equal to x to the r.

00:05 So next we need to find y prime, which is equal to r times x to the r minus 1.

00:09 Y double prime is equal to r times r minus 1, minus 1, minus 1 times x to the r minus 2.

00:16 And then y triple prime is going to be r times r minus 1 times r minus 2 times x to the r minus 3.

00:24 So substituting these into our differential equation here, we have x cubed times r minus 1, times r minus 2 times x to the r minus 3 is r minus 3 plus 3 x squared r times r minus 1 times x to the r minus 2 minus 6x sorry that should be 6x here okay 6x times r times x to the r minus 1 is equal to 0.

00:57 So next we'll simplify this sum.

01:01 X cubed and x r minus 3 just be becomes x to the r.

01:04 So we have left, we have r cubed minus three r squared, minus two, or sorry, plus two are here.

01:15 And then times x to the r, and then plus this x squared can combine with this x to the r minus two, just become x to the r.

01:24 So we have three times r squared plus, oh sorry, minus three r, x to the r, and then minus six.

01:36 Again, this x and this x to the r minus 1 combined to become just x to the r.

01:41 So we just get 6r, x to the r is equal to 0.

01:46 Okay? so next we can factor on an x of the r from all three of these terms.

01:50 So you get x to the r...

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