Q7E - Fundamentals Of Differential Equations And Boundary Value Problems

Question

Find a general solution to the Cauchy-Euler equation ${\mathit{x}}^{\mathbf{3}}{\mathit{y}}^{\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{\text{'}}}\mathbf{-}\mathbf{3}{\mathit{x}}^{\mathbf{2}}{\mathit{y}}^{\mathbf{\text{'}}\mathbf{\text{'}}}\mathbf{+}\mathbf{6}\mathit{x}{\mathit{y}}^{\mathbf{\text{'}}}\mathbf{-}\mathbf{6}\mathit{y}\mathbf{=}{\mathit{x}}^{\mathbf{-}\mathbf{1}}\mathbf{,}\mathbf{ }\mathbf{ }\mathbf{ }\mathit{x}\mathbf{>}\mathbf{0}\mathbf{,}$

given that $\left\{x,x^2,x^3\right\}$is a fundamental solution set for the corresponding homogeneous equation

Step-by-Step Solution

Answer

The general solution is  $y=C_{1}x+C_2{x}^2+C_3{x}^3-\frac{1}{24x}$

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 complementary solution

It is given that $x^3{y}^{\text{'}\text{'}\text{'}}-3x^2{y}^{\text{'}\text{'}}+6x{y}^{\text{'}}-6y=\frac{1}{x}$

Also,$\left\{x,x^2,x^3\right\}$ is the fundamental solution

The complementary solution is as follows:

$y_c=C_{1}x+C_2{x}^2+C_3{x}^3$

Now, find out the particular solution of the given differential equation.

Step 3: Calculate Wornkians

Calculate the Wronskian of the above set as follows:

$$\begin{gathered} W\left[x,x^2,x^3\right]=\left|\begin{array}{ccc}x& x^2& x^3\\ 1& 2x& 3x^2\\ 0& 2& 6x\end{array}\right| \\ =12x^3-6x^3-6x^3+2x^3 \\ =2x^3 \end{gathered}$$

The value of wronkians $W_1,W_2,W_3$ is:

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

Step 4: For particular solution

Evaluate the particular solution 

$$\begin{gathered} y^{\text{'}\text{'}\text{'}}-\frac{3}{x}{y}^{\text{'}\text{'}}+\frac{6y^{\text{'}}}{x^2}-\frac{6y}{x^3}=\frac{1}{x^4} \\ {y}_p=x\int \frac{x^4\left(\frac{1}{x^4}\right)}{\left(\frac{x^4}{x^4}\right)2x^3}dx+x^2\int \frac{-2x^3\left(\frac{1}{x^4}\right)}{\left(\frac{x^4}{x^4}\right)2x^3}+x^3\int \frac{x^2\left(\frac{1}{x^4}\right)}{2x^3}dx \\ {y}_p=\frac{-x}{4x^2}+x^2\int \frac{-2}{2x^4}dx+x^3\int \frac{dx}{2x^5} \end{gathered}$$