I have the following question:
Given the Euler-Cauchy equation
$$x^2\frac{d^2y}{dx^2}-3x\frac{dy}{dx}-5y=x^5$$
Solve the associated homogeneous equation by considering a solution of the form $y(x) = x^r$ where $r$ is a constant.
To obtain a particular solution consider solutions of the form $y(x)=ax^s\ln(x)$ for suitable constants $a$ and $s$.
Now I know how to solve the inhomogenous part, and found:
$$y_h=Ax^5+Bx^{-1}.$$
But I have no idea how to go about finding the particular solution by considering solutions of 'that form'.
Any hints are greatly appreciated.
$\endgroup$ 01 Answer
$\begingroup$Don't know about the form $y(x)=ax^s\ln(x)$, but you can apply Variation of Parameters to find out the Particular solution.
The given Euler-Cauchy equation can be modified as:$$\frac{d^2y}{dx^2}-\frac{3}{x}\cdot\frac{dy}{dx}-\frac{5y}{x^2}=x^3$$
The general Homogeneous solution is: $$y_h=Ax^5+Bx^{-1}$$
Let, the particular solution for the same is: $$y_p=C(x)\cdot x^5+D(x)\cdot x^{-1}$$
Therefore, $$C'(x)\cdot x^5+D'(x)\cdot x^{-1}=0\qquad\cdots(i)\\\text{and, }\quad5C'(x)\cdot x^4-D'(x)\cdot\frac{1}{x^2}=x^3\qquad\cdots(ii)$$ From (i) and (ii) we can write, $C'(x)=\dfrac{1}{6x}\implies C(x)=\dfrac{1}{6}\ln|x|$; and $D'(x)=-\dfrac{x^5}{6}\implies D(x)=-\dfrac{1}{36}x^6$.
Therefore Particular Solution: $$y_p=\dfrac{1}{6}x^5\ln|x|-\dfrac{1}{36}x^5.$$ So, $$y=y_h+y_p=Ax^5+Bx^{-1}+\dfrac{1}{6}x^5\ln|x|-\dfrac{1}{36}x^5.$$
$\endgroup$ 2
