Q10E - Fundamentals Of Differential Equations And Boundary Value Problems

Question

Given that $\left\{x,x^{-1},x^4\right\}$  is a fundamental solution set for the homogeneous equation corresponding to the equation${\mathit{x}}^{\mathbf{3}}{\mathit{y}}^{\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{\text{'}}}\mathbf{-}{\mathit{x}}^{\mathbf{2}}{\mathit{y}}^{\mathbf{\text{'}}\mathbf{\text{'}}}\mathbf{-}\mathbf{4}\mathit{x}{\mathit{y}}^{\mathbf{\text{'}}}\mathbf{+}\mathbf{4}\mathit{y}\mathbf{=}\mathit{g}\mathbf{\left(}\mathit{x}\mathbf{\right)}\mathbf{,}\mathbf{ }\mathbf{ }\mathbf{ }\mathit{x}\mathbf{>}\mathbf{0}\mathbf{,}$ determine a formula involving integrals for a particular solution.

Step-by-Step Solution

Answer

The particular solution is $y_p=\frac{-x}{6}\int \frac{gdx}{x^2}+\frac{1}{10x}\int gdx+\frac{x^4}{15}\int \frac{gdx}{x^5}$

Step 1: Definition

Variation of parameters, also known as variation of constants, is a general method to solve inhomogeneous linear ordinary differential equations.

Step 2: Find Wronkians

Consider the differential equation $x^3{y}^{\text{'}\text{'}\text{'}}-x^2{y}^{\text{'}\text{'}}-4x{y}^{\text{'}}+4y=g$

The fundamental set of solution is $\left\{x,x^{-1},x^4\right\}$

$$\begin{gathered} W\left[\begin{array}{ccc}x& x^{-1}& x^4\end{array}\right]=\left|\begin{array}{ccc}x& \frac{1}{x}& x^4\\ 1& \frac{-1}{x^2}& 4x^3\\ 0& \frac{2}{x^3}& 12x^2\end{array}\right| \\ =x(-12-8)-(12x-2x) \\ =-30x \end{gathered}$$

Step 3: Wronkians

The value of Wronkians $W_1,W_2,W_3$ is given by:

$$\begin{gathered} W_1=(-1{)}^{3-1}W\left[\begin{array}{cc}{x}^{-1}& x^4\end{array}\right] \\ =\left|\begin{array}{cc}\frac{1}{x}& x^4\\ \frac{-1}{x^2}& 4x^3\end{array}\right| \\ =5x^2 \\ {W}_2=(-1{)}^{3-2}W\left[\begin{array}{cc}x& x^4\end{array}\right] \\ =-3x^4 \\ {W}_3=(-1{)}^{3-3}W\left[\begin{array}{cc}x& x^{-1}\end{array}\right] \\ =\left|\begin{array}{cc}x& x^{-1}\\ 1& \frac{{\displaystyle -1}}{{\displaystyle x^2}}\end{array}\right| \\ =\frac{-2}{x} \end{gathered}$$

The particular solution is

$$\begin{gathered} y_p=x\int \frac{5x^2\left(\frac{g}{x^3}\right)dx}{-30x}+\frac{1}{x}\int \frac{-3x^4\times \left(\frac{g}{x^3}\right)dx}{-30x}+x^4\int \frac{\frac{-2}{x}\left(\frac{g}{x^3}\right)dx}{-30x} \\ {y}_p=\frac{-x}{6}\int \frac{gdx}{x^2}+\frac{1}{10x}\int gdx+\frac{x^4}{15}\int \frac{gdx}{x^5} \end{gathered}$$