I'm trying to solve the following differential equation:
$ (x^{2} -1) y'' - (4x^{2} -3x-5) y' + (4x^{2} -6x -5) y = e^{2x}$
By the method of factorizing operators, so I rewrite the differential equations like this:
$ [(x^{2} -1) D^2 - (4x^{2} -3x-5) D + (4x^{2} -6x -5)] y= e^{2x}$
So I'm basically trying to factorize the expression:
$ [(x^{2} -1) D^2 - (4x^{2} -3x-5) D + (4x^{2} -6x -5)]$
So I think this really just falls under the category of factorizing a quadratic with variable coefficients, once factorized, the rest of the procedure is non problematic, so can anyone give me a hint on how to factorize this?
$$ (x^{2} -1) y'' - (4x^{2} -3x-5) y' + (4x^{2} -6x -5) y = e^{2x} \tag 1$$ This is an inhomogeneous linear ODE.
Change of function : $y(x)=e^{2x}u(x)\quad;\quad y'=e^{2x}(2u+u')\quad;\quad y''=e^{2x}(4u+4u'+u'')$
$ (x^{2} -1) e^{2x}(4u+4u'+u'') - (4x^{2} -3x-5) e^{2x}(2u+u')+ (4x^{2} -6x -5) e^{2x}u = e^{2x}$ $$ (x^2-1)u'' + ( 3x+1)u'+ u = 1 \tag 2$$ $$\big((x^2-1)u'\big)'+\big((x+1)u\big)'=1$$ $$(x^2-1)u'+(x+1)u=x+c_1$$ There is no difficulty to solve this first order linear ODE. The result is : $$u=\frac{x}{x-1}+c_1\frac{\ln|x+1|}{x-1}+c_2\frac{1}{x-1}$$ $$y(x)=\frac{e^{2x}}{x-1}\big(x+c_1\ln|x+1|+c_2\big)$$