Solve: $(2x+3)^2\frac{d^2y}{dx^2}-(2x+3)\frac{dy}{dx}-12y=6x$

Question

Solve: $(2x+3)^2\frac{d^2y}{dx^2}-(2x+3)\frac{dy}{dx}-12y=6x$

Is my equation (2) correct? Because while solving the auxiliary equation, I am not getting a 'neat' value for $m$.

Answer 1

You ca use this substitution as well:

Take $2x+3=t$ to get a Cauchy-Euler non homogeneous linear ODE: $$4t^2y''-2ty'-12y=6x$$

Try this!

Answer 2

You can use Euler 's equation like @Resident has pointed out

You can also substitute $$y=(2x+3)^m$$ Then solve the equation $$2m^2-3m-6=0$$ $$\implies m=\frac {3 \pm \sqrt {57}}4$$ The general solution of the homogeneous equation is $$y_h(x)=c_1(2x+3)^{\frac {3 + \sqrt {57}}4}+c_2(2x+3)^{\frac {3 - \sqrt {57}}4}$$

the particular solution is $$y_p=ax+b, y'_p=a, y''_p=0$$ $$-a(2x+3)-12(ax+b)=6x$$ $$\implies y_p=-\frac 37x+\frac 3 {28}$$