Question

Use variation of parameters to find a particular solution, given the solutions \(y_{1}, y_{2}\) of the complementary equation. $$ x^{2} y^{\prime \prime}-4 x y^{\prime}+6 y=x^{5 / 2}, x>0 ; \quad y_{1}=x^{2}, y_{2}=x^{3} $$

Short answer

#tag_title#Step 2: Calculate the integrals for the multipliers #tag_content#We will now calculate the integrals for the multipliers of $y_1$ and $y_2$ using the Wronskian: Let $v_1(x)$ and $v_2(x)$ be two functions such that: $$ y_p = v_1(x)y_1(x) + v_2(x)y_2(x) $$ Then, the first and second derivatives of $y_p$ are: $$ y_p' = v_1'y_1 + v_1y_1' + v_2'y_2 + v_2y_2' $$ $$ y_p'' = v_1''y_1 + 2v_1'y_1' + v_1y_1'' + v_2''y_2 + 2v_2'y_2' + v_2y_2'' $$ Now we plug in the values of $y_1, y_1', y_1'', y_2, y_2'$, and $y_2''$ into the given equation $x^2y'' - 4xy' + 6y = x^{\frac{5}{2}}$, which simplifies to: $$ v_1' = \frac{x^{\frac{1}{2}}}{2x^4} $$ $$ v_2' = \frac{x^{\frac{1}{2}}}{6x^5} $$ Next, we will integrate both functions separately with respect to $x$: $$ v_1 = \int \frac{x^{\frac{1}{2}}}{2x^4} dx $$ $$ v_2 = \int \frac{x^{\frac{1}{2}}}{6x^5} dx $$ Upon integrating and simplifying, we get: $$ v_1 = -\frac{1}{6x^{\frac{3}{2}}} + C_1 $$ $$ v_2 = \frac{1}{20x^{\frac{3}{2}}} + C_2 $$ However, for the particular solution, we may choose to ignore the constants of integration $C_1$ and $C_2$. Thus, we have: $$ v_1 = -\frac{1}{6x^{\frac{3}{2}}} $$ $$ v_2 = \frac{1}{20x^{\frac{3}{2}}} $$ #tag_title#Step 3: Write out the particular solution #tag_content#Now, we will plug in the values of $v_1$, $v_2$, $y_1$, and $y_2$ into our particular solution: $$ y_p = v_1(x)y_1(x) + v_2(x)y_2(x) $$ Therefore, the particular solution is: $$ y_p = \left(-\frac{1}{6x^{\frac{3}{2}}}\right)(x^2) + \left(\frac{1}{20x^{\frac{3}{2}}}\right)(x^3) $$ #tag_title#Step 4: Simplify the particular solution #tag_content#Finally, we will simplify our particular solution: $$ y_p = -\frac{1}{6x^{\frac{1}{2}}} + \frac{1}{20x^{\frac{1}{2}}} $$ Thus, the particular solution for the given second-order differential equation is: $$ y_p = \left(\frac{1}{20x^{\frac{1}{2}}} - \frac{1}{6x^{\frac{1}{2}}}\right) $$

Step-by-step solution

Write down the given equations and Wronskian

Given the second-order differential equation: $$ x^2y'' - 4xy' + 6y = x^{\frac{5}{2}}, \quad x > 0 $$ And the solutions of the complementary equation: $$ y_1 = x^2, \quad y_2 = x^3 $$ We will first find the Wronskian (\(W\)) of \(y_1\) and \(y_2\): $$ W = \begin{vmatrix} y_1 & y_2\\ y_1' & y_2' \end{vmatrix} $$

Key concepts

Differential Equations

A differential equation is a mathematical equation that relates some function with its derivatives. In applications, the functions generally represent physical quantities, the derivatives represent their rates of change, and the equation defines a relationship between the two.

For the exercise at hand, we're dealing with a second-order differential equation, where the highest derivative is a second derivative, denoted as 'y''', which signifies the rate of change of the rate of change of y. In the field of differential equations, there are various methods to solve these intricate relationships. One such method is the variation of parameters, which is a technique used to find a particular solution to nonhomogeneous differential equations like the one presented.

Wronskian

The Wronskian is a determinant used in the study of differential equations to check whether two functions are linearly independent solutions of the equation. Named after the Polish mathematician Józef Hoene-Wroński, the Wronskian, in the context of our problem, involves functions that are solutions to the homogeneous part of our original equation.

By calculating the determinant of the matrix composed of these solutions and their first derivatives, we obtain a value that, if non-zero, confirms the linear independence of the functions. This is a key step in applying the method of variation of parameters. Here's how the Wronskian is calculated for our problem's solutions, \(y_1 = x^2\) and \(y_2 = x^3\):

\[ W = \begin{vmatrix} y_1 & y_2 \ y_1' & y_2' \end{vmatrix} \]
Where \(y_1'\) and \(y_2'\) are the first derivatives of \(y_1\) and \(y_2\), respectively.

Complementary Equation

The complementary equation, sometimes called the homogeneous equation, is a differential equation that equals zero. For nonhomogeneous differential equations, like the one in our exercise, we have terms on both sides of the equation. However, the homogeneous or complementary part of it lacks these additional terms and is crucial in investigating the behavior of a differential system.

The solutions found from the complementary equation, referred to as the complementary functions or the homogeneous solutions, are essential in determining the structure of the general solution. In our case, \(y_1 = x^2\) and \(y_2 = x^3\) are two solutions to the complementary equation. They will later assist us in finding the particular solution via variation of parameters.

Particular Solution

In differential equations, a particular solution is a single, specific solution to a nonhomogeneous differential equation. It satisfies not only the differential equation but also fits specific initial conditions or boundary values.

After establishing the general solution using the solutions to the complementary equation, one can superimpose a particular solution to accommodate the nonhomogeneous part of the equation. In the method of variation of parameters used in our exercise, the particular solution is calculated by adjusting parameters within the complementary functions, ensuring that the proposed solution satisfies the original nonhomogeneous equation. In simpler terms, this method helps us plug in specific functions into the original equation to obtain a new solution that includes the nonhomogenous term, in this scenario, \(x^{5/2}\).