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. ${\mathit{x}}^{\mathbf{2}}\mathit{y}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{-}\mathbf{3}\mathit{x}\mathit{y}\mathbf{\text{'}}\mathbf{+}\mathbf{13}\mathit{y}\mathbf{=}\mathbf{4}\mathbf{+}\mathbf{3}\mathit{x}$

Short answer

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

$y\left(x\right)=k_1{x}^2\cos \left(6\ln x\right)+k_2{x}^2\sin \left(6\ln x\right)+\frac{3}{10}x+\frac{4}{13}$

Step-by-step solution

Deriving the given solution by the differential equation

We have the Cauchy Euler equation:

$x^2\frac{d^{2}y}{d{x}^2}-3x\frac{dy}{dx}+13y=4+3x$

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

$\begin{array}{rcl}{x}^2\times \frac{1}{x^2}\left(\frac{d^{2}y}{d{t}^2}-\frac{dy}{dt}\right)-3x\times \frac{1}{x}\frac{dy}{dt}+13y& =& 4+3e^t\\ \frac{d^{2}y}{d{t}^2}-\frac{dy}{dt}-3\frac{dy}{dt}+13y& =& 4+3e^t\\ \frac{d^{2}y}{d{t}^2}-4\frac{dy}{dt}+13y& =& 4+3e^t\end{array}$

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

$\begin{array}{rcl}{m}^2{e}^{mt}-4m{e}^{mt}+13e^{mt}& =& 0\\ \left(m^2-4m+13\right)e^{mt}& =& 0\\ & & \end{array}$

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

$m^2-4m+13=0$

Then the roots are$m_{1,2}=\frac{4\pm \sqrt{16-4\times 13}}{2}=\frac{4\pm \sqrt{-36}}{2}=2\pm 6i$ which are conjugate complex.

Solving homogeneous solution using Cauchy Euler equation

We can obtain the corresponding homogeneous solution as

$\begin{array}{rcl}{y}_c& =& c_1{e}^{(2-6i)t}+c_{2}t{e}^{(2-6i)t}\\ & =& c_1{e}^{2t}{e}^{6it}+c_2{e}^{2t}{e}^{6it}\\ & =& c_1{e}^{2t}(\cos 6t+\sin 6t)+c_2{e}^{2t}(\cos 6t-\sin 6t)\\ & =& \left(c_1+c_2\right)e^{2t}\cos 6t+\left(c_1-c_2\right)e^{2t}\sin 6t\\ & =& k_1{e}^{2t}\cos 6t+k_2{e}^{2t}\sin 6t\\ & & \end{array}$

Third, now we have to obtain the particular solution by assuming that $y_p=A{e}^t+B$since we have$g\left(t\right)=4+3e^t$ , then differentiate with respect to t, then we have

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

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

$$\begin{gathered} A{e}^t-4A{e}^t+13\left(A{e}^t+B\right)=4+3e^t \\ 10A{e}^t+13B=4+3e^t \end{gathered}$$

$10A=3then:A=\frac{3}{10}$

$and13B=4then:B=\frac{4}{13}$

Then we can obtain the particular solution for the given Cauchy Euler equation as $y_p=\frac{3}{10}{e}^t+\frac{4}{13}$

Deriving the given solution by the differential equation

Solving the equation, we get

$\begin{array}{rcl}y\left(x\right)& =& y_c+y_p\\ & =& k_1{e}^{2t}\cos 6t+k_2{e}^{2t}\sin 6t+\frac{3}{10}{e}^t+\frac{4}{13}\\ & & \end{array}$

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

$y\left(x\right)=k_1{e}^{2\ln x}\cos \left(6\ln x\right)+k_2{e}^{2\ln x}\sin \left(6\ln x\right)+\frac{3}{10}{e}^{\ln x}+\frac{4}{13}$

$$\begin{gathered} =k_1{e}^{\ln x^2}\cos \left(6\ln x\right)+k_2{e}^{\ln x^2}\sin \left(6\ln x\right)+\frac{3}{10}{e}^{\ln x}+\frac{4}{13} \\ =k_1{x}^2\cos \left(6\ln x\right)+k_2{x}^2\sin \left(6\ln x\right)+\frac{3}{10}x+\frac{4}{13} \end{gathered}$$

The solution of the given Cauchy Euler equation $x^2\frac{d^{2}y}{d{x}^2}-3x\frac{dy}{dx}+13y=4+3x$.