The given equation is
$$x^2y^{''}+xy^{'}+y=\ln(x)$$
Finding the Aux. Equation :
$a=1, b=1, c=1 $
$$am^2+(b-a)m+c=0$$
$$m^2+1=0$$
Knowing that this specific type of equation results in the $m$ values being $\pm i$.
Now using the formulas below we can substitute in
$$y=x^\alpha C_1\cos(\beta\ln(x))+C_2\sin(\beta\ln(x)) \space (Equ. 1)$$
$$m=\alpha+\beta i$$
$$\alpha=0 ,\space \beta=\pm 1$$
And my final answer is:
$$y=C_1\cos(\ln(x))+C_2\sin(\ln(x))$$
There are a few places in this problem I have questions about:
In Equ. 1 the $x^{\alpha}$ term, that exponent is not the same $a$ variable that comes form the Aux. Equ., correct?
Is my $\beta$ value $\pm 1$ or $\pm i$?
Also is my answer correct, meaning that my answer is the general solution to the given? If not, am I close to it or are there further calculations to be made?
Best Answer
In Equ. 1 the $x^α$ term, that exponent is not the same a variable that comes form the Aux. Equ., correct?
Well it's the same $\alpha=0$ since you have $m^2+1=0 \implies m=0\pm i$. so it's $x^{\alpha}=x^0=1$. For the coeffcient $\beta$ it's equal to $1$ and not $\pm i$. The solution to the homogeneous equation is:$y=c_1\cos (\ln x)+c_2\sin (\ln x)$
In fact you can write the solution as: $y=c_1x^{\alpha+i \beta}+c_2x^{\alpha-i \beta}$ where $m=\alpha \pm i \beta$ is the solution to the polynomial characteristic.Then apply Euler's formula $e^{ix}=\cos x +i \sin x$.
$$x^{\alpha+i \beta}=x^{\alpha }x^{i \beta}=x^{\alpha }(e^{i \beta \ln x})=x^{\alpha }(\cos (\beta \ln x)+i \sin (\beta \ln x) )$$