reduction of order when one solution is given

Question

I'm currently working on this differential equation, where one solution is already known, however, I'm having difficulties:

The equation:

$(x+1)^2y''-4(x+1)y'+6y=0$

where:

$y_1{(x})=(x+1)^2;$ where $y_2{(x)}=u_xy_1{(x)}$

Then differentiating to obtain $y_2'$and $y_2''$, I get:

$u_x(x+1)^2+2u_x'(x+1)+2u_x''$

substituting this in the differential equation:

$2u_x''(x+1)^2-8u_x'(x+1)^2+6u(x+1)^2=0$

when expanding the bracket:

$2u_x''x^2+4u_x''x+2u_x''-8u_x'x^2-16u_x'x-8u_x'+6ux^2+12ux+6u=0$

I'm stuck on what to do next?

Answer 1

$$(x+1)^2y''-4(x+1)y'+6y=0$$ For simplicity set $z=x+1$ $$z^2y''-4zy'+6y=0$$ Then $y=z^2 v$ $$z^2(v'z^2+2zv)'-4z(v'z^2+2zv)+6z^2v=0$$ $$z^2(v''z^2+4zv'+2v)-4v'z^3-2z^2v=0$$ $$z^2(v''z^2+4zv')-4v'z^3=0$$ $$v''=0$$ The order is reduced. $$v=c_1z+c_2$$ $$y=vz^2=c_1z^3+c_2z^2$$ $$y(x)=c_1(x+1)^3+c_2(x+1)^2$$

Answer 2

$$z^22y''−4zy'+6y=0,~~z=x+1,$$ is an Euler-Cauchy equation with characteristic polynomial $$ 0=r(r-1)-4r+6=(r-2)(r-3) $$ This gives $(z^2,z^3)$ or $\bigl((x+1)^2,(x+1)^3\bigr)$ as basis solutions.

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In the reduction approach, it pays to be careful in the derivatives computation. With $y_2(x)=u(x)(x+1)^2$ the derivatives are $$ y_2'(x)=(x+1)^2u'(x)+2(x+1)u(x)\\ y_2''(x)=(x+1)^2u''(x)+4(x+1)u'(x)+2u $$ where you can see that you reverted somehow the place of the derivatives.

In general, the terms without derivative in $u$ have to cancel, else you have an error in the prior computations.

Answer 3

Let $t:=x+1$. We solve

$$t^2y''-4ty'+6y=0,$$ knowing the solution $y_1=t^2$.

With $y_2=vy_1$,

$$t^2(v''t^2+4v't+2v)-4t(v't^2+2vt)+6vt^2=v''t^4=0$$

and

$$v(t)=at+b.$$

Hence,

$$y(t)=(cx+d)(x+1)^2.$$