Undetermined coefficients is great, right?
We love it!
Except for when it's useless.
Consider an example...
When Undetermined Coefficients Fails
Suppose we need to to solve the equation $$ y''+9y=\tan(3x) $$ Question #1: What particular solution might work here?
Since $\tan(3x)$ doesn't have derivatives and integrals that look similar to itself (like $\cos (ax),$ $\sin (ax),$ $e^{ax}$
and polynomials $a_nx^n+\cdots+a_x+a_0$), understanding what to guess for a particular solution $y_p$ becomes a great deal more complicated.
Question #2: Is there any hope?
Question #2: Is there any hope?
Answer to Question #2
Yes!
Today we learn a method for finding particular solutions that relies considerably less on guesswork.
The method is called variation of parameters.
The Big Idea
For a second-order linear equation, $$ y''+p(x)y'+q(x)y=f(x) $$ we assume that we have two linearly independent solutions $y_1$ and $y_2$ to the homogeneous equation $$ y''+p(x)y'+q(x)y=0 $$ We take the homogeneous solution $$ y=C_1y_1+C_2y_2 $$ We then allow the parameters $C_1$ and $C_2$ vary to obtain a particular solution which has the form $$ y_p=v_1(x)y_1+v_2(x)y_2 $$
Variation of Parameters
Through a lot of tedious calculation based upon the above assumptions, we may solve for $v_1'(x)$ and $v_2'(x)$ by solving the system of equations $$ \begin{cases} y_1v_1'+y_2 v_2'&=0\\ y_1'v_1'+y_2' v_2'&=f(x)\\ \end{cases} $$ In matrix form we may also write $$ \left[ \begin{array}{cc} y_1 & y_2\\ y_1' & y_2'\\ \end{array} \right] \left[ \begin{array}{c} v_1' \\ v_2' \\ \end{array} \right] = \left[ \begin{array}{c} 0 \\ f(x) \\ \end{array} \right] $$ Once we have $v_1'$ and $v_2',$ we may integrate to get $v_1$ and $v_2$ to obtain our particular solution $$ y_p=v_1y_1+v_2y_2 $$
Example
Find a particular and general solution to the equation $$ y''+9y=\tan(3x) $$
We shall use the variation of parameters technique which requires knowing the
the general solution to the homogeneous equation.
Solving $y''+9y=0,$ the characteristic equation is $r^2+9=0$ so that $r=\pm 3i.$ Then $$ y_h=C_1\cos(3x)+C_2\sin(3x) $$ solves the homogeneous equation.
We now "vary parameters" and suppose a particular solution has the form $$ y_p=v_1(x)\cos(3x)+v_2(x)\sin(3x) $$ Then $v_1$ and $v_2$ must satisfy $$ \begin{cases} y_1v_1'+y_2 v_2'&=0\\ y_1'v_1'+y_2' v_2'&=f(x)\\ \end{cases} $$ or $$ \begin{cases} \cos(3x)v_1'+\sin(3x) v_2'&=0\\ -3\sin(3x)v_1'+3\cos(3x)v_2'&=\tan(3x)\\ \end{cases} $$ Multiplying the top equation by $3\tan(3x),$ we have $3\sin(3x)v_1'+3\tan(3x)\sin(3x)v_2'=0.$
Adding this equation to the second equation in the above system gives $$ 3\cos(3x)v_2'+3\tan(3x)\sin(3x)v_2'=\tan(3x) $$ so that $$ \begin{array}{lll} \displaystyle v_2'&\displaystyle=\frac{\tan(3x)}{3\cos(3x)+3\tan(3x)\sin(3x)} &\mbox{}\\ \displaystyle &\displaystyle=\frac{\tan(3x)}{3\cos(3x)+3\tan(3x)\sin(3x)}\color{magenta}{\frac{\cos(3x)}{\cos(3x)}} &\mbox{}\\ \displaystyle &\displaystyle=\frac{\sin(3x)}{3\cos^2(3x)+3\sin^2(3x)}&\mbox{}\\ \displaystyle &\displaystyle=\frac{1}{3}\sin(3x)&\mbox{}\\ \end{array} $$ Then from the first equation of the system $$ \begin{array}{lll} &\displaystyle \cos(3x)v_1'+\sin(3x) v_2'=0&\mbox{}\\ \implies &\displaystyle \cos(3x)v_1'+\sin(3x)\left(\frac{1}{3}\sin(3x)\right)=0&\mbox{}\\ \implies &\displaystyle \cos(3x)v_1'+\frac{1}{3}\sin^2(3x)=0&\mbox{}\\ \implies &\displaystyle v_1'=-\frac{1}{3}\frac{\sin^2(3x)}{\cos(3x)}&\mbox{}\\ \end{array} $$ The solution to the system is $$ v_1'=\frac{1}{3}\frac{\sin^2(3x)}{\cos(3x)} \,\,\,\,\mbox{and}\,\,\,\, v_2'=\frac{1}{3}\sin(3x) $$ We then have $$ \begin{array}{lll} \displaystyle v_1&\displaystyle=\int -\frac{1}{3}\frac{\sin^2(3x)}{\cos(3x)} \, dx &\mbox{}\\ \displaystyle &\displaystyle=-\frac{1}{3}\int \frac{1-\cos^2(3x)}{\cos(3x)} \, dx &\mbox{}\\ \displaystyle &\displaystyle=-\frac{1}{3}\int \sec(3x)-\cos(3x) \, dx &\mbox{}\\ \displaystyle &\displaystyle=-\frac{1}{3}\left(\frac{1}{3}\ln|\sec(3x)+\tan(3x)|-\frac{1}{3}\sin(3x)\right) &\mbox{}\\ \displaystyle &\displaystyle=-\frac{1}{9}\ln|\sec(3x)+\tan(3x)|+\frac{1}{9}\sin(3x) &\mbox{}\\ \end{array} $$ and $$ \begin{array}{lll} \displaystyle v_2&\displaystyle= \int \frac{1}{3}\sin(3x) \, dx &\mbox{}\\ \displaystyle &\displaystyle= -\frac{1}{9}\cos(3x)&\mbox{}\\ \end{array} $$ So, our particular solution is $$ \begin{array}{ll} y_p&= \displaystyle v_1(x)\cos(3x)+v_2(x)\sin(3x)\\ &= \displaystyle \left(-\frac{1}{9}\ln|\sec(3x)+\tan(3x)|+\frac{1}{9}\sin(3x)\right)\cos(3x)-\frac{1}{9}\cos(3x)\sin(3x)\\ &= \displaystyle -\frac{1}{9}\cos(3x)\ln|\sec(3x)+\tan(3x)|+\frac{1}{9}\sin(3x)\cos(3x)-\frac{1}{9}\cos(3x)\sin(3x)\\ &= \displaystyle -\frac{1}{9}\cos(3x)\ln|\sec(3x)+\tan(3x)|\\ \end{array} $$ The general solution to the equation is then $$ y=y_h+y_p=C_1\cos(3x)+C_2\sin(3x)-\frac{1}{9}\cos(3x)\ln|\sec(3x)+\tan(3x)| $$
Solving $y''+9y=0,$ the characteristic equation is $r^2+9=0$ so that $r=\pm 3i.$ Then $$ y_h=C_1\cos(3x)+C_2\sin(3x) $$ solves the homogeneous equation.
We now "vary parameters" and suppose a particular solution has the form $$ y_p=v_1(x)\cos(3x)+v_2(x)\sin(3x) $$ Then $v_1$ and $v_2$ must satisfy $$ \begin{cases} y_1v_1'+y_2 v_2'&=0\\ y_1'v_1'+y_2' v_2'&=f(x)\\ \end{cases} $$ or $$ \begin{cases} \cos(3x)v_1'+\sin(3x) v_2'&=0\\ -3\sin(3x)v_1'+3\cos(3x)v_2'&=\tan(3x)\\ \end{cases} $$ Multiplying the top equation by $3\tan(3x),$ we have $3\sin(3x)v_1'+3\tan(3x)\sin(3x)v_2'=0.$
Adding this equation to the second equation in the above system gives $$ 3\cos(3x)v_2'+3\tan(3x)\sin(3x)v_2'=\tan(3x) $$ so that $$ \begin{array}{lll} \displaystyle v_2'&\displaystyle=\frac{\tan(3x)}{3\cos(3x)+3\tan(3x)\sin(3x)} &\mbox{}\\ \displaystyle &\displaystyle=\frac{\tan(3x)}{3\cos(3x)+3\tan(3x)\sin(3x)}\color{magenta}{\frac{\cos(3x)}{\cos(3x)}} &\mbox{}\\ \displaystyle &\displaystyle=\frac{\sin(3x)}{3\cos^2(3x)+3\sin^2(3x)}&\mbox{}\\ \displaystyle &\displaystyle=\frac{1}{3}\sin(3x)&\mbox{}\\ \end{array} $$ Then from the first equation of the system $$ \begin{array}{lll} &\displaystyle \cos(3x)v_1'+\sin(3x) v_2'=0&\mbox{}\\ \implies &\displaystyle \cos(3x)v_1'+\sin(3x)\left(\frac{1}{3}\sin(3x)\right)=0&\mbox{}\\ \implies &\displaystyle \cos(3x)v_1'+\frac{1}{3}\sin^2(3x)=0&\mbox{}\\ \implies &\displaystyle v_1'=-\frac{1}{3}\frac{\sin^2(3x)}{\cos(3x)}&\mbox{}\\ \end{array} $$ The solution to the system is $$ v_1'=\frac{1}{3}\frac{\sin^2(3x)}{\cos(3x)} \,\,\,\,\mbox{and}\,\,\,\, v_2'=\frac{1}{3}\sin(3x) $$ We then have $$ \begin{array}{lll} \displaystyle v_1&\displaystyle=\int -\frac{1}{3}\frac{\sin^2(3x)}{\cos(3x)} \, dx &\mbox{}\\ \displaystyle &\displaystyle=-\frac{1}{3}\int \frac{1-\cos^2(3x)}{\cos(3x)} \, dx &\mbox{}\\ \displaystyle &\displaystyle=-\frac{1}{3}\int \sec(3x)-\cos(3x) \, dx &\mbox{}\\ \displaystyle &\displaystyle=-\frac{1}{3}\left(\frac{1}{3}\ln|\sec(3x)+\tan(3x)|-\frac{1}{3}\sin(3x)\right) &\mbox{}\\ \displaystyle &\displaystyle=-\frac{1}{9}\ln|\sec(3x)+\tan(3x)|+\frac{1}{9}\sin(3x) &\mbox{}\\ \end{array} $$ and $$ \begin{array}{lll} \displaystyle v_2&\displaystyle= \int \frac{1}{3}\sin(3x) \, dx &\mbox{}\\ \displaystyle &\displaystyle= -\frac{1}{9}\cos(3x)&\mbox{}\\ \end{array} $$ So, our particular solution is $$ \begin{array}{ll} y_p&= \displaystyle v_1(x)\cos(3x)+v_2(x)\sin(3x)\\ &= \displaystyle \left(-\frac{1}{9}\ln|\sec(3x)+\tan(3x)|+\frac{1}{9}\sin(3x)\right)\cos(3x)-\frac{1}{9}\cos(3x)\sin(3x)\\ &= \displaystyle -\frac{1}{9}\cos(3x)\ln|\sec(3x)+\tan(3x)|+\frac{1}{9}\sin(3x)\cos(3x)-\frac{1}{9}\cos(3x)\sin(3x)\\ &= \displaystyle -\frac{1}{9}\cos(3x)\ln|\sec(3x)+\tan(3x)|\\ \end{array} $$ The general solution to the equation is then $$ y=y_h+y_p=C_1\cos(3x)+C_2\sin(3x)-\frac{1}{9}\cos(3x)\ln|\sec(3x)+\tan(3x)| $$
Variation of Parameters Formula
Using Cramer's Rule, we can solve $$ \left[ \begin{array}{cc} y_1 & y_2\\ y_1' & y_2'\\ \end{array} \right] \left[ \begin{array}{c} v_1' \\ v_2' \\ \end{array} \right] = \left[ \begin{array}{c} 0 \\ f(x) \\ \end{array} \right] $$ for $v_1'$ and $v_2'$ straightaway as $$ v_1'=\frac{\left|\begin{array}{cc} 0 & y_2 \\f(x) & y_2'\\ \end{array}\right|}{\left|\begin{array}{cc} y_1&y_2\\y_1'&y_2'\\ \end{array}\right|} =-\frac{y_2f(x)}{W[y_1,y_2]} =-\frac{y_2f(x)}{y_1y_2'-y_2y_1'} \\\\ v_2'=\frac{\left|\begin{array}{cc} y_1 & 0 \\y_1' & f(x)\\ \end{array}\right|}{\left|\begin{array}{cc} y_1&y_2\\y_1'&y_2'\\ \end{array}\right|} =\frac{y_1f(x)}{W[y_1,y_2]} =\frac{y_1f(x)}{y_1y_2'-y_2y_1'} $$
Variation of Parameters Formula
Thus, from the above, we have the following formulas. $$ v_1=-\int \frac{y_2f(x)}{W[y_1,y_2]}\, dx=-\int \frac{y_2f(x)}{y_1y_2'-y_2y_1'}\, dx \\\\ v_2=\int \frac{y_1f(x)}{W[y_1,y_2]} \, dx=\int \frac{y_1f(x)}{y_1y_2'-y_2y_1'}\, dx $$
Example
Use the above formulas to obtain $v_1$ and $v_2$ in the particular solution $y_p$ for the equation $$ y''+9y=\tan(3x) $$
Since $y_1=\cos(3x)$ and $y_2=\sin(3x),$ we have $y_1'=-3\sin(3x)$ and $y_2'=3\cos(3x).$
Then $$ \begin{array}{lll} W[y_1,y_2]&=\left|\begin{array}{cc} y_1&y_2\\y_1'&y_2'\\ \end{array}\right|\\ &=y_1y_2'-y_2y_1'\\ &=\cos(3x)(3\cos(3x))-(\sin(3x))(-3\sin(3x))\\ &=3\cos^2(3x)+3\sin^2(3x)\\ &=3\\ \end{array} $$ so that $$ \begin{array}{lll} \displaystyle v_1&\displaystyle=-\int \frac{y_2f(x)}{W[y_1,y_2]}\, dx &\mbox{}\\ \displaystyle &\displaystyle=-\int \frac{\sin(3x)\tan(3x)}{3}\, dx &\mbox{}\\ \displaystyle &\displaystyle=-\frac{1}{3}\int \frac{\sin^2(3x)}{\cos(3x)}\, dx &\mbox{}\\ \displaystyle &\displaystyle=-\frac{1}{3}\int \frac{1-\cos^2(3x)}{\cos(3x)} \, dx &\mbox{}\\ \displaystyle &\displaystyle=-\frac{1}{3}\int \sec(3x)-\cos(3x) \, dx &\mbox{}\\ \displaystyle &\displaystyle=-\frac{1}{3}\left(\frac{1}{3}\ln|\sec(3x)+\tan(3x)|-\frac{1}{3}\sin(3x)\right) &\mbox{}\\ \displaystyle &\displaystyle=-\frac{1}{9}\ln|\sec(3x)+\tan(3x)|+\frac{1}{9}\sin(3x) &\mbox{}\\ \end{array} $$ and $$ \begin{array}{lll} \displaystyle v_2&\displaystyle=\int \frac{y_1f(x)}{W[y_1,y_2]} \, dx &\mbox{}\\ \displaystyle &\displaystyle=\int \frac{\cos(3x)\tan(3x)}{3} \, dx &\mbox{}\\ \displaystyle &\displaystyle=\frac{1}{3}\int \sin(3x) \, dx &\mbox{}\\ \displaystyle &\displaystyle=-\frac{1}{9}\cos(3x) &\mbox{}\\ \end{array} $$
Then $$ \begin{array}{lll} W[y_1,y_2]&=\left|\begin{array}{cc} y_1&y_2\\y_1'&y_2'\\ \end{array}\right|\\ &=y_1y_2'-y_2y_1'\\ &=\cos(3x)(3\cos(3x))-(\sin(3x))(-3\sin(3x))\\ &=3\cos^2(3x)+3\sin^2(3x)\\ &=3\\ \end{array} $$ so that $$ \begin{array}{lll} \displaystyle v_1&\displaystyle=-\int \frac{y_2f(x)}{W[y_1,y_2]}\, dx &\mbox{}\\ \displaystyle &\displaystyle=-\int \frac{\sin(3x)\tan(3x)}{3}\, dx &\mbox{}\\ \displaystyle &\displaystyle=-\frac{1}{3}\int \frac{\sin^2(3x)}{\cos(3x)}\, dx &\mbox{}\\ \displaystyle &\displaystyle=-\frac{1}{3}\int \frac{1-\cos^2(3x)}{\cos(3x)} \, dx &\mbox{}\\ \displaystyle &\displaystyle=-\frac{1}{3}\int \sec(3x)-\cos(3x) \, dx &\mbox{}\\ \displaystyle &\displaystyle=-\frac{1}{3}\left(\frac{1}{3}\ln|\sec(3x)+\tan(3x)|-\frac{1}{3}\sin(3x)\right) &\mbox{}\\ \displaystyle &\displaystyle=-\frac{1}{9}\ln|\sec(3x)+\tan(3x)|+\frac{1}{9}\sin(3x) &\mbox{}\\ \end{array} $$ and $$ \begin{array}{lll} \displaystyle v_2&\displaystyle=\int \frac{y_1f(x)}{W[y_1,y_2]} \, dx &\mbox{}\\ \displaystyle &\displaystyle=\int \frac{\cos(3x)\tan(3x)}{3} \, dx &\mbox{}\\ \displaystyle &\displaystyle=\frac{1}{3}\int \sin(3x) \, dx &\mbox{}\\ \displaystyle &\displaystyle=-\frac{1}{9}\cos(3x) &\mbox{}\\ \end{array} $$
Bonus Example with Variable-Coefficients
Given that $\displaystyle y_1=\frac{1}{x-1}$ and $\displaystyle y_2=\frac{1}{x+1}$ are solutions to the corresponding homogeneous equation to $$ (x^2-1)y''+4xy'+2y=-\frac{2}{x+1} $$ find a particular and general solution to the full equation.
We shall use the variation of parameters formula to to this situation.
We first put the equation in standard form. $$ y''+\frac{4x}{x^2-1}y'+\frac{2}{x^2-1}y=-\frac{2}{(x+1)(x^2-1)} $$ Then for this situation $$ f(x)=-\frac{2}{(x+1)(x^2-1)}=-\frac{2}{(x+1)^2(x-1)} $$ Now, $\displaystyle y_1'=-\frac{1}{(x-1)^{2}}$ and $\displaystyle y_2'=-\frac{1}{(x+1)^{2}}.$
The Wronskian is then $$ \begin{array}{ll} W[y_1,y_2]&=\displaystyle\left|\begin{array}{cc} y_1&y_2\\y_1'&y_2'\\ \end{array}\right|\\ &=\displaystyle\left|\begin{array}{cc} (x-1)^{-1}&(x+1)^{-1}\\-(x-1)^{-2}&-(x+1)^{-2}\\ \end{array}\right|\\ &=\displaystyle -(x-1)^{-1}(x+1)^{-2}+(x+1)^{-1}(x-1)^{-2}\\ &=\displaystyle -\frac{1}{x-1}\frac{1}{(x+1)^{2}}+\frac{1}{(x+1)}\frac{1}{(x-1)^{2}}\\ &=\displaystyle -\frac{x-1}{(x-1)^2(x+1)^{2}}+\frac{x+1}{(x+1)^2(x-1)^{2}}\\ &=\displaystyle \frac{2}{(x+1)^2(x-1)^{2}}\\ \end{array} $$ We then have $$ \begin{array}{lll} \displaystyle v_1&\displaystyle=-\int \frac{y_2f(x)}{W[y_1,y_2]}\, dx &\mbox{}\\ \displaystyle &\displaystyle=-\int \frac{1}{x+1}\left(-\frac{2}{(x+1)^2(x-1)}\right)\frac{(x+1)^2(x-1)^{2}}{2} \, dx &\mbox{}\\ \displaystyle &\displaystyle=\int \frac{x-1}{x+1} \, dx &\mbox{}\\ \displaystyle &\displaystyle=\int \frac{u-2}{u} \, du &\mbox{$u=x+1$ and $du=dx$}\\ \displaystyle &\displaystyle=\int 1-\frac{2}{u} \, du &\mbox{}\\ \displaystyle &\displaystyle=u-2\ln |u| &\mbox{}\\ \displaystyle &\displaystyle=x+1-2\ln |x+1| &\mbox{}\\ \end{array} $$ For simplicity, we may take $v_1=x-2\ln|x+1|.$
For $v_2,$ $$ \begin{array}{lll} \displaystyle v_2&\displaystyle=\int \frac{y_1f(x)}{W[y_1,y_2]}\, dx &\mbox{}\\ \displaystyle &\displaystyle=\int \frac{1}{x-1}\left(-\frac{2}{(x+1)^2(x-1)}\right)\frac{(x+1)^2(x-1)^{2}}{2} \, dx &\mbox{}\\ \displaystyle &\displaystyle=\int -1 \, dx &\mbox{}\\ \displaystyle &\displaystyle=-x &\mbox{}\\ \end{array} $$ The particular solution is then $$ \begin{array}{lll} \displaystyle y_p&\displaystyle=v_1y_1+v_2y_2 &\mbox{}\\ \displaystyle &\displaystyle= \left(x-2\ln|x+1|\right)\frac{1}{x-1} +(-x)\frac{1}{x+1} &\mbox{}\\ \displaystyle &\displaystyle= \frac{x-2\ln|x+1|}{x-1} -\frac{x}{x+1} &\mbox{}\\ \end{array} $$ The general solution is $$ y=y_h+y_p=\frac{C_1}{x-1}+\frac{C_2}{x+1}+\frac{x-2\ln|x+1|}{x-1} -\frac{x}{x+1} $$
We first put the equation in standard form. $$ y''+\frac{4x}{x^2-1}y'+\frac{2}{x^2-1}y=-\frac{2}{(x+1)(x^2-1)} $$ Then for this situation $$ f(x)=-\frac{2}{(x+1)(x^2-1)}=-\frac{2}{(x+1)^2(x-1)} $$ Now, $\displaystyle y_1'=-\frac{1}{(x-1)^{2}}$ and $\displaystyle y_2'=-\frac{1}{(x+1)^{2}}.$
The Wronskian is then $$ \begin{array}{ll} W[y_1,y_2]&=\displaystyle\left|\begin{array}{cc} y_1&y_2\\y_1'&y_2'\\ \end{array}\right|\\ &=\displaystyle\left|\begin{array}{cc} (x-1)^{-1}&(x+1)^{-1}\\-(x-1)^{-2}&-(x+1)^{-2}\\ \end{array}\right|\\ &=\displaystyle -(x-1)^{-1}(x+1)^{-2}+(x+1)^{-1}(x-1)^{-2}\\ &=\displaystyle -\frac{1}{x-1}\frac{1}{(x+1)^{2}}+\frac{1}{(x+1)}\frac{1}{(x-1)^{2}}\\ &=\displaystyle -\frac{x-1}{(x-1)^2(x+1)^{2}}+\frac{x+1}{(x+1)^2(x-1)^{2}}\\ &=\displaystyle \frac{2}{(x+1)^2(x-1)^{2}}\\ \end{array} $$ We then have $$ \begin{array}{lll} \displaystyle v_1&\displaystyle=-\int \frac{y_2f(x)}{W[y_1,y_2]}\, dx &\mbox{}\\ \displaystyle &\displaystyle=-\int \frac{1}{x+1}\left(-\frac{2}{(x+1)^2(x-1)}\right)\frac{(x+1)^2(x-1)^{2}}{2} \, dx &\mbox{}\\ \displaystyle &\displaystyle=\int \frac{x-1}{x+1} \, dx &\mbox{}\\ \displaystyle &\displaystyle=\int \frac{u-2}{u} \, du &\mbox{$u=x+1$ and $du=dx$}\\ \displaystyle &\displaystyle=\int 1-\frac{2}{u} \, du &\mbox{}\\ \displaystyle &\displaystyle=u-2\ln |u| &\mbox{}\\ \displaystyle &\displaystyle=x+1-2\ln |x+1| &\mbox{}\\ \end{array} $$ For simplicity, we may take $v_1=x-2\ln|x+1|.$
For $v_2,$ $$ \begin{array}{lll} \displaystyle v_2&\displaystyle=\int \frac{y_1f(x)}{W[y_1,y_2]}\, dx &\mbox{}\\ \displaystyle &\displaystyle=\int \frac{1}{x-1}\left(-\frac{2}{(x+1)^2(x-1)}\right)\frac{(x+1)^2(x-1)^{2}}{2} \, dx &\mbox{}\\ \displaystyle &\displaystyle=\int -1 \, dx &\mbox{}\\ \displaystyle &\displaystyle=-x &\mbox{}\\ \end{array} $$ The particular solution is then $$ \begin{array}{lll} \displaystyle y_p&\displaystyle=v_1y_1+v_2y_2 &\mbox{}\\ \displaystyle &\displaystyle= \left(x-2\ln|x+1|\right)\frac{1}{x-1} +(-x)\frac{1}{x+1} &\mbox{}\\ \displaystyle &\displaystyle= \frac{x-2\ln|x+1|}{x-1} -\frac{x}{x+1} &\mbox{}\\ \end{array} $$ The general solution is $$ y=y_h+y_p=\frac{C_1}{x-1}+\frac{C_2}{x+1}+\frac{x-2\ln|x+1|}{x-1} -\frac{x}{x+1} $$