use the substitution ( t=ln(x) ) to transform the given cauchy - euler equation to a differential equation…

Explanation:

Step1: Substitute (t = \ln(x))

We know that (x = e^{t}), (y'=\frac{dy}{dx}=\frac{dy}{dt}\cdot\frac{dt}{dx}=\frac{1}{x}\frac{dy}{dt}), and (y''=\frac{d^{2}y}{dx^{2}}=\frac{d}{dx}(\frac{1}{x}\frac{dy}{dt})=-\frac{1}{x^{2}}\frac{dy}{dt}+\frac{1}{x^{2}}\frac{d^{2}y}{dt^{2}})

Substitute into the equation (x^{2}y''-3xy'+13y = 6 + 5x)

[ \begin{align*} x^{2}\left(-\frac{1}{x^{2}}\frac{dy}{dt}+\frac{1}{x^{2}}\frac{d^{2}y}{dt^{2}}\right)-3x\left(\frac{1}{x}\frac{dy}{dt}\right)+13y&=6 + 5e^{t}\ \frac{d^{2}y}{dt^{2}}-4\frac{dy}{dt}+13y&=6 + 5e^{t} \end{align*} ]

Step2: Solve the homogeneous equation

The homogeneous equation is (\frac{d^{2}y}{dt^{2}}-4\frac{dy}{dt}+13y = 0)

The characteristic equation is (r^{2}-4r + 13=0)

Using the quadratic formula (r=\frac{4\pm\sqrt{16 - 52}}{2}=2\pm3i)

The complementary solution (y_{c}(t)=e^{2t}(C_{1}\cos(3t)+C_{2}\sin(3t)))

Step3: Find a particular solution

For the non - homogeneous part (6+5e^{t})

For (y_{p1}): Let (y_{p1}=A), then (y_{p1}''-4y_{p1}'+13y_{p1}=13A = 6\Rightarrow A=\frac{6}{13})

For (y_{p2}): Let (y_{p2}=Be^{t}), then (y_{p2}''-4y_{p2}'+13y_{p2}=(B - 4B+13B)e^{t}=10Be^{t})

Since (10Be^{t}=5e^{t}\Rightarrow B=\frac{1}{2})

The particular solution (y_{p}(t)=\frac{6}{13}+\frac{1}{2}e^{t})

Step4: Write the general solution in terms of (t)

(y(t)=e^{2t}(C_{1}\cos(3t)+C_{2}\sin(3t))+\frac{6}{13}+\frac{1}{2}e^{t})

Step5: Substitute back (t=\ln(x))

Since (e^{t}=x) and (e^{2t}=x^{2}), (t = \ln(x))

(y(x)=x^{2}(C_{1}\cos(3\ln(x))+C_{2}\sin(3\ln(x)))+\frac{6}{13}+\frac{1}{2}x)

Answer:

(y(x)=x^{2}(C_{1}\cos(3\ln(x))+C_{2}\sin(3\ln(x)))+\frac{6}{13}+\frac{1}{2}x)