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.${\mathit{x}}^{\mathbf{2}}\mathit{y}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{+}\mathbf{10}\mathit{x}\mathit{y}\mathbf{\text{'}}\mathbf{+}\mathbf{8}\mathit{y}\mathbf{=}{\mathit{x}}^{\mathbf{2}}$

Short answer

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

$y\left(x\right)=c_1{x}^{-1}+c_2{x}^{-8}+\frac{1}{30}{x}^2$

Step-by-step solution

Deriving the given solution by the differential equation

We have the Cauchy Euler equation

$x^2\frac{d^{2}y}{d{x}^2}+10x\frac{dy}{dx}+8y=x^2$

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:

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

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

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 (c) and (e) into equation (1), then we have

$$\begin{gathered} x^2\times \frac{1}{x^2}\left(\frac{d^{2}y}{d{t}^2}-\frac{dy}{dt}\right)+10x\times \frac{1}{x}\frac{dy}{dt}+8y=e^{2t} \\ \frac{d^{2}y}{d{t}^2}-\frac{dy}{dt}+10\frac{dy}{dt}+8y=e^{2t} \\ \frac{d^{2}y}{d{t}^2}+9\frac{dy}{dt}+8y=e^{2t} \end{gathered}$$

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

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

After that, substitute with $y=e^{mt}$and equations (3) and (4) into equation (2), then we obtain

$$\begin{gathered} m^2{e}^{mt}+9m{e}^{mt}+8e^{mt}=0 \\ \left(m^2+9m+8\right)e^{mt}=0 \end{gathered}$$

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

$$\begin{gathered} m^2+9m+8=0 \\ (m+1)(m+8)=0 \end{gathered}$$

Then the roots are $m_1=-1and{m}_2=-8$which are real and distinct.

Solving homogeneous solution using Cauchy Euler equation

We can obtain the corresponding homogeneous solution as

$\begin{array}{rcl}{y}_c& =& c_1{e}^{m_{1}t}+c_{2}t{e}^{m_{2}t}\\ & =& c_1{e}^{-t}+c_2{e}^{8t}\end{array}$

Third, now we have to obtain the particular solution by assuming that $y_p=A{e}^{2t}$since we have$g\left(t\right)=e^{2t}$ , then differentiate with respect to t, then we have

$\frac{d{y}_p}{dt}=2A{e}^{2t}$

$\frac{d^2{y}_p}{d{t}^2}=4A{e}^{2t}$

After, that, substitute with$y_p=A{e}^{2t}$ and from equations (f) and (g) into equation (2), then we have

$\begin{array}{rcl}4A{e}^{2t}+9\times 2A{e}^{2t}+8A{e}^{2t}& =& e^{2t}\\ (4A+18A+8A)e^{2t}& =& e^{2t}\\ 30A{e}^{2t}& =& e^{2t}\\ 30A& =& 1\\ A& =& \frac{1}{30}\\ & & \end{array}$

Then we can obtain the particular solution for the given Cauchy Euler equation as$y_p=\frac{1}{30}{e}^{2t}$

Deriving the given solution by the differential equation

Solving the equation, we get

$\begin{array}{rcl}y\left(x\right)& =& y_c+y_p\\ & =& c_1{e}^{-t}+c_2{e}^{-8t}+\frac{1}{30}{e}^{2t}\end{array}$

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

$\begin{array}{rcl}y\left(x\right)& =& c_1{e}^{-\ln x}+c_2{e}^{-8\ln x}+\frac{1}{30}{e}^{2\ln x}\\ & =& c_1{e}^{\ln x^{-1}}+c_2{e}^{\ln x^{-8}}+\frac{1}{30}{e}^{\ln x^2}\\ & =& c_1{x}^{-1}+c_2{x}^{-8}+\frac{1}{30}{x}^2\\ & & \end{array}$

The solution of the given Cauchy Euler equation $x^2\frac{d^{2}y}{d{x}^2}+10x\frac{dy}{dx}+8y=x^2$.