Q8E - Fundamentals Of Differential Equations And Boundary Value Problems

Question

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

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

Step-by-Step Solution

Answer

The general solution is  $y\left(x\right)=C_{1}x+C_{2}x\ln x+C_3{x}^3-x^2$

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   $x^3{y}^{\text{'}\text{'}\text{'}}-2x^2{y}^{\text{'}\text{'}}+3x{y}^{\text{'}}-3y=0$    -------(1)

The fundamental solution set is $\left\{x,x\ln x,x^3\right\}$

So, the complementary function is $y_c\left(x\right)=C_{1}x+C_{2}x\ln x+C_3{x}^3$

Step 3: Calculate Wornkians

Find $W,W_k,k=1,2,3$as follows:

$$\begin{gathered} W\left[x,x\ln x,x^3\right]=\left|\begin{array}{ccc}x& x\ln x& x^3\\ 1& 1+\ln x& 3x^2\\ 0& \frac{1}{x}& 6x\end{array}\right| \\ =4x^2 \end{gathered}$$

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

Step 4: Calculate V i

Evaluate.

$$\begin{gathered} v_1\left(x\right)=\int \frac{g\left(x\right)W_1}{W}dx \\ =\int \frac{\frac{1}{x}\left(2x^3\ln x-x^3\right)}{4x^2}dx \\ =\frac{1}{4}(2x\ln x-3x) \end{gathered}$$

$$\begin{gathered} v_2\left(x\right)=\int \frac{g\left(x\right)W_2}{W}dx \\ =\int \frac{\frac{1}{x}\left(-2x^3\right)}{4x^2}dx \\ =-\frac{x}{2} \\ {v}_3\left(x\right)=\int \frac{g\left(x\right)W_3}{W}dx \\ =-\frac{1}{4x} \end{gathered}$$

Step 5: Particular solution

Since $\left\{x,x\ln x,x^3\right\}$ is a fundamental solution set, so a particular solution of the form,

$y_p\left(x\right)=v_1\left(x\right)x+v_2\left(x\right)x\ln x+v_3\left(x\right)x^3$

The particular solution is,

$$\begin{gathered} y_p\left(x\right)=v_1\left(x\right)x+v_2\left(x\right)x\ln x+v_3\left(x\right)x^3 \\ =\frac{1}{4}(2x\ln x-3x)·x+\left(-\frac{x}{2}\right)·x\ln x+\left(-\frac{1}{4x}\right)·x^3 \\ =\frac{1}{2}{x}^2\ln x-\frac{3}{4}{x}^2-\frac{1}{2}{x}^2\ln x-\frac{1}{4}{x}^2 \\ =-x^2 \end{gathered}$$

Thus, the general solution of the equation (1) is,

$$\begin{gathered} y\left(x\right)=y_c\left(x\right)+y_p\left(x\right) \\ =C_{1}x+C_{2}x\ln x+C_3{x}^3-x^2 \end{gathered}$$