If you solve$(7.17)$when$f\left(x\right)=0$by assuming a solution$y=x^k$, show that the quadratic equation for$k$is the same as the auxiliary equation for the$z$equation (7.20). Thus show (see Section 5) that if the two values of$k$are equal, the second solution is not a power of$k$but is$x^{k}lnx$. Also show that if$k$is complex, say$k=a\pm bi$, the solutions are$x^{a}cos(blnx)$and$x^{a}sin(blnx)$or other equivalent forms [see$\left(5.16\right)$to$\left(5.18\right)$].

Consider the given equation.

$a_2{x}^2\frac{d^{2}y}{d{x}^2}+a_{1}x\frac{dy}{dx}+a_{0}y=f\left(x\right)\dots \left(1\right)$

Auxiliary equation is an algebraic equation of degree $n$upon which depends the solution of a given$n^{th}$-order differential equation or difference equation.

It’s one of the solutions is,

$y=x^k$

So,

$$\begin{gathered} \frac{dy}{dx}=k{x}^{k-1} \\ \frac{d^{2}y}{d{x}^2}=k(k-1)x^{k-2} \end{gathered}$$

Thus,

$$\begin{gathered} a_{2}k(k-1)+a_{1}k+a_0=0 \\ {a}_2{k}^2+\left(a_1-a_2\right)k+a_0=0 \end{gathered}$$

It is same as the auxiliary of equation (1).

If the masses $k_1$and$k_2$ are equal. Then the solution is,

$$\begin{gathered} y=\left(c_{1+}{c}_{2}z\right)e^{k_{1}z} \\ =(c_{1+}{c}_2\log x)x^{k_1} \end{gathered}$$

So, $x^{k_1}$and $x^{k_1}$$\log x$are the two solutions of the equation.

Now, $k_1=k_2$. So, the other solution is,

$y=x^{k}logx$

If$k$ is complex such that,

$k=a\pm ib$

The solution of the equation is,

$$\begin{gathered} y=e^{az}(c_1\cos bz+c_2\sin bz) \\ =x^a(c_1\cos b\log x+c_2\sin b\log x) \end{gathered}$$

So, the solution for the given equation is$x^{a}cosblogx$ and $x^{a}sinblogx$.

Therefore, the proof that if the two values of $k$are equal then the second is$x^{k}lnx$ and if $k$is complex then the solutions are $x^{a}cos(blnx)$and $\sin (b\ln x)$is stated above.