Question

Use the substitution $\mathit{x}\mathbf{=}{\mathit{e}}^{\mathbf{t}}$to transform the given Cauchy-Euler equation to a differential equation with constant coefficients. Solve the original equation by solving the new equation using the procedures in Sections 4.3-4.5.${\mathit{x}}^{\mathbf{3}}\mathit{y}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{-}\mathbf{3}{\mathit{x}}^{\mathbf{2}}\mathit{y}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{+}\mathbf{6}\mathit{x}\mathit{y}\mathbf{\text{'}}\mathbf{-}\mathbf{6}\mathit{y}\mathbf{=}\mathbf{3}\mathbf{+}\mathit{l}\mathit{n}{\mathit{x}}^{\mathbf{3}}$

Short answer

The solution of the Cauchy-Euler equation to a differential equation with constant coefficients is

$y\left(x\right)=c_{1}x+c_2{x}^2+c_3{x}^3-\frac{1}{2}\ln x+\frac{17}{12}$

Step-by-step solution

Deriving the given solution by a differential equation

We have the Cauchy Euler equation

$x^3\frac{d^{3}y}{d{x}^3}-3x^2\frac{2}{d{x}^2}+6x\frac{dy}{dx}-6y=3+\ln x^3$

We have to solve it as the following technique:

First, use the substitution $x=e^t$and make expansion then make a partial differentiation as the following:

$$\begin{gathered} x=e^t \\ t=\ln x \\ dt=\frac{1}{x}dx \\ \frac{dt}{dx}=\frac{1}{x} \end{gathered}$$

Using the chain rule,

$\frac{dy}{dx}=\frac{dy}{dt}\times \frac{dt}{dx}$

Substitute from equation (a) into equation (b), then we have

$\frac{dy}{dx}=\frac{1}{x}\frac{dy}{dt}$

After that, differentiate equation (c) with respect to x, then we have

$\begin{array}{rcl}\frac{d^{2}y}{d{x}^2}& =& \frac{1}{x}\times \frac{d}{dx}\left(\frac{dy}{dt}\right)+\frac{dy}{dt}\times \frac{d}{dx}\left(\frac{1}{x}\right)\\ & =& \frac{1}{x}\frac{dy}{dx}\left(\frac{dy}{dt}\right)-\frac{1}{x^2}\frac{dy}{dt}\end{array}$

After that, substitute from equation (c) into equation (d), then we have

$\begin{array}{rcl}\frac{d^{2}y}{d{x}^2}& =& \frac{1}{x}\frac{dy}{dt}\left(\frac{1}{x}\frac{dy}{dt}\right)-\frac{1}{x^2}\frac{dy}{dt}\\ & =& \frac{1}{x^2}\frac{d^{2}y}{d{t}^2}-\frac{1}{x^2}\frac{dy}{dt}\\ & =& \frac{1}{x^2}\left(\frac{d^{2}y}{d{t}^2}-\frac{dy}{dt}\right)\\ & & \\ & & \end{array}$

Deriving the given solution by the differential equation

After that, substitute from equations (e) with respect to x, then we have

$\begin{array}{rcl}\frac{d^{3}y}{d{x}^3}& =& \frac{1}{x^2}\times \frac{1}{x}\frac{dy}{dt}\left(\frac{d^{2}y}{d{t}^2}-\frac{dy}{dt}\right)+\frac{2}{x^3}\left(\frac{dy}{dt}-\frac{d^{2}y}{d{t}^2}\right)\\ & =& \frac{1}{x^3}\left(\frac{d^{3}y}{d{t}^3}-\frac{d^{2}y}{d{t}^2}\right)+\frac{2}{x^3}\left(\frac{dy}{dt}-\frac{d^{2}y}{d{t}^2}\right)\\ & =& \frac{1}{x^3}\left(\frac{d^{3}y}{d{t}^3}-3\frac{d^{2}y}{d{t}^2}+2\frac{dy}{dt}\right)\\ & & \end{array}$

After that, substitute from equation (c) into equation (f), then we have

$\begin{array}{rcl}\frac{d^{3}y}{d{x}^3}& =& \frac{1}{x^2}\times \frac{1}{x}\frac{dy}{dt}\left(\frac{d^{2}y}{d{t}^2}-\frac{dy}{dt}\right)+\frac{2}{x^3}\left(\frac{dy}{dt}-\frac{d^{2}y}{d{t}^2}\right)\\ & =& \frac{1}{x^3}\left(\frac{d^{3}y}{d{t}^3}-\frac{d^{2}y}{d{t}^2}\right)+\frac{2}{x^3}\left(\frac{dy}{dt}-\frac{d^{2}y}{d{t}^2}\right)\\ & =& \frac{1}{x^3}\left(\frac{d^{3}y}{d{t}^3}-3\frac{d^{2}y}{d{t}^2}+2\frac{dy}{dt}\right)\\ & & \end{array}$

After that, substitute with $x=e^t$and from equations (c),(e), and (g) into equation (1), then we have

$$\begin{gathered} x^3\times \frac{1}{x^3}\left(\frac{d^{3}y}{d{t}^3}-3\frac{d^{2}y}{d{t}^2}+2\frac{dy}{dt}\right)-3x^2\times \frac{1}{x^2}\left(\frac{d^{2}y}{d{t}^2}-\frac{dy}{dt}\right)+6x\times \frac{1}{x}\frac{dy}{dt}-6y=3+\ln e^{3t} \\ \frac{d^{3}y}{d{t}^3}-3\frac{d^{2}y}{d{t}^2}+2\frac{dy}{dt}-3\frac{d^{2}y}{d{t}^2}+3\frac{dy}{dt}+6\frac{dy}{dt}-6y=3+3t \\ \frac{d^{3}y}{d{t}^3}-6\frac{d^{2}y}{d{t}^2}+11\frac{dy}{dt}-6y=3+3t \end{gathered}$$

Second, now we have to the solution for the corresponding homogeneous differential equation:

$\frac{d^{3}y}{d{t}^3}-6\frac{d^{2}y}{d{t}^2}+11\frac{dy}{dt}-6y=0$

Now, we have to solve this second order differential equation by assuming that$y=e^{mt}$ is a solution for this D.E, then differentiate with respect to t, then we have

$\frac{dy}{dt}=m{e}^{mt}$

Differentiate another time with respect to t, then we have

$\frac{d^{2}y}{d{t}^2}=m^2{e}^{mt}$and $\frac{d^{3}y}{d{t}^3}=m{e}^3{e}^{mt}$

After that, substitute with $y=A{e}^{mt}$and equations (4),(5), and (6) into equation (3), then we obtain

$$\begin{gathered} m^3{e}^{mt}-6m^2{e}^{mt}+11m{e}^{mt}-6e^{mt}=0 \\ \left(m^3-6m^2+11m-6\right)e^{mt}=0 \end{gathered}$$

Since $e^{mt}$cannot be equal to 0, then we have the auxiliary solution as

$$\begin{gathered} m^3-6m^2+11m-6=0 \\ (m-1)\left(m^2-5m+6\right)=0 \\ (m-1)(m-2)(m-3)=0 \end{gathered}$$

Then the roots are$m_1=1,m_2=2and{m}_3=3$ which are real and distinct.

Solving homogeneous solution using Cauchy Euler equation

We can obtain the corresponding homogeneous solution as

$\begin{array}{rcl}{y}_c& =& c_1{e}^{m_{1}t}+c_2{e}^{m_{2}t}+c_3{e}^{m_{3}t}\\ & =& c_1{e}^t+c_2{e}^{2t}+c_3{e}^{3t}\end{array}$

Third, now we have to obtain the particular solution by assuming that $y_p=At+B$since we have $g\left(t\right)=3t+3$, then differentiate with respect to t, then we have

$$\begin{gathered} \frac{d{y}_p}{dt}=A \\ \frac{d^2{y}_p}{d{t}^2}=0 \\ \frac{d^3{y}_p}{d{t}^3}=0 \end{gathered}$$

After, that, substitute with $y_p=A{t}^2+Bt+C$and from equations (h), (i), and (j) into equation (2), then we have

$$\begin{gathered} 0-0+11A-6(At+B)=3+3t \\ (11A-6B)-6At=3+3t \end{gathered}$$

$-10A=3then:A=-\frac{1}{2}$

$and11A-6B=3then:B=\frac{17}{12}$

Then we can obtain the particular solution for the given Cauchy Euler equation as$y_p=-\frac{1}{2}t+\frac{17}{12}$

Deriving the given solution by the differential equation

Solving the equation, we get

$\begin{array}{rcl}y\left(x\right)& =& y_c+y_p\\ & =& c_1{e}^t+c_2{e}^{2t}+c_3{e}^{3t}-\frac{1}{2}t+\frac{17}{12}\end{array}$

Finally, we have to back substitute with$t=\ln x$ , then we have

$\begin{array}{rcl}y\left(x\right)& =& c_1{e}^{\ln x}+c_2{e}^{2\ln x}+c_3{e}^{3\ln x}-\frac{1}{2}\ln x+\frac{17}{12}\\ & =& c_1{e}^{\ln x}+c_2{e}^{\ln x^2}+c_3{e}^{\ln x^3}-\frac{1}{2}\ln x+\frac{17}{12}\\ & =& c_{1}x+c_2{x}^2+c_3{x}^3-\frac{1}{2}\ln x+\frac{17}{12}\\ & & \end{array}$

The solution of the given Cauchy Euler equation$x^3\frac{d^{3}y}{d{x}^3}-3x^2\frac{2}{d{x}^2}+6x\frac{dy}{dx}-$$6y=3+\ln x^3$