4.3.2.13 Solved By transformation on $B\left(x\right)$ for ODE $A{y}^{\prime \prime }\left(x\right)+B{y}^{\prime }\left(x\right)+C\left(x\right)y\left(x\right)=0$

ode internal name "second_order_ode_non_constant_coeff_transformation_on_B"

This method is tried to reduce the order ode the ODE by one, by doing direct transformation on $B\left(x\right)$ for the ode

$$A\left(x\right)y^{\prime \prime }\left(x\right)+B\left(x\right)y^{\prime }\left(x\right)+C\left(x\right)y\left(x\right)=0$$

Let

$$y=Bv$$

Then $y^{\prime }=B^{\prime }v+v^{\prime }B$ and $y^{\prime \prime }=B^{\prime \prime }v+B^{\prime }{v}^{\prime }+v^{\prime \prime }B+v^{\prime }{B}^{\prime }=v^{\prime \prime }B+2v^{\prime }{B}^{\prime }+B^{\prime \prime }v$ then the original ode becomes

$$\begin{array}{rl}A(v^{\prime \prime }B+2v^{\prime }{B}^{\prime }+B^{\prime \prime }v)+B(B^{\prime }v+v^{\prime }B)+CBv& =0\\ AB{v}^{\prime \prime }+(2A{B}^{\prime }+B^2)v^{\prime }+(A{B}^{\prime \prime }+B{B}^{\prime }+CB)v& =0\end{array}$$

Now we check if $A{B}^{\prime \prime }+B{B}^{\prime }+CB=0$ or not. If it is zero, then this method works and we can now solve

$$AB{v}^{\prime \prime }+(2A{B}^{\prime }+B^2)v^{\prime }=0$$

Using $u=v^{\prime }$ which reduces the order to one.

$$AB{u}^{\prime }+(2A{B}^{\prime }+B^2)u=0$$

This is first order ode now. Solved for $u$ gives $v^{\prime }$ which is solved for $v$ as first order ode. Then $y=Bv$ and we are done. This method only works of course if $A{B}^{\prime \prime }+B{B}^{\prime }+CB=0$ comes out to be zero. Here is an example

4.3.2.13.1 Example 1

$$x{y}^{\prime \prime }+(-1-x)y^{\prime }+y=0$$

Here $A=x,B=(-1-x)$ and $C=1$, hence $B^{\prime }=-1,B^{\prime \prime }=0$ and therefore

$$\begin{array}{rl}A{B}^{\prime \prime }+B{B}^{\prime }+CB& =0+(-1-x)(-1)+(-1-x)\\ & =1+x-1-x\\ & =0\end{array}$$

It works. Hence the reduces ode becomes

$$AB{v}^{\prime \prime }+(2A{B}^{\prime }+B^2)v^{\prime }=0$$

Let $u=v^{\prime }$ then

$$\begin{array}{rl}AB{u}^{\prime }+(2A{B}^{\prime }+B^2)u& =0\\ x\left((-1-x)\right)u^{\prime }+(-2x+{(-1-x)}^2)u& =0\\ u-x{u}^{\prime }+u{x}^2-x^2{u}^{\prime }& =0\\ u^{\prime }(-x-x^2)+u(1+x^2)& =0\\ u^{\prime }-\frac{(1+x^2)}{(x+x^2)}u& =0\end{array}$$

This is linear first order ode solved using integrating factor which gives

$$u=c_1\frac{x{e}^x}{{(1+x)}^2}$$

Hence since $v^{\prime }=u$ then

$$v^{\prime }=c_1\frac{x{e}^x}{{(1+x)}^2}$$

This is quadrature. Solving gives

$$v=c_2+c_1\frac{e^x}{1+x}$$

Therefore

$$\begin{array}{rl}y& =Bv\\ & =(-1-x)(c_2+c_1\frac{e^x}{1+x})\\ & =c_2(1+x)+c_1{e}^x\end{array}$$

Note that this method is sensitive to the ODE is written. If we divide the ode by $A$ is becomes

$$y^{\prime \prime }+\frac{(-1-x)}{x}+\frac{1}{x}y=0$$

And now $A=1,B=\frac{(-1-x)}{x}$ and $C=\frac{1}{x}$, hence $B^{\prime }=-\frac{1}{x}+\frac{1+x}{x^2}$ and $B^{\prime \prime }=\frac{2}{x^2}-\frac{2}{x^3}(1+x)$ then

$$\begin{array}{rl}A{B}^{\prime \prime }+B{B}^{\prime }+CB& =(\frac{2}{x^2}-\frac{2}{x^3}(1+x))+\left(\frac{(-1-x)}{x}\right)(-\frac{1}{x}+\frac{1+x}{x^2})+\frac{1}{x}\left(\frac{(-1-x)}{x}\right)\\ & =-\frac{1}{x^3}(x^2+2x+3)\\ & \ne 0\end{array}$$

So this method now fails to reduce the ode order by one. So in practice, I try first on the ode as given, and then try again by normalizing it so that $B$ is not rational function and try again. In other words, given an ode $y^{\prime \prime }+\frac{(-1-x)}{x}+\frac{1}{x}y=0$ then try with $B=\frac{(-1-x)}{x}$ and if this fails, try again after multiplying the ode by $x$ so now $B=(-1-x)$ and $A=x$ and $C=1$ and see if this works or not. This method of course only works when $B$ is not zero.