Find the general solution of (x^2-1) y'' + 4x y' + 2y = 6x…

Find the general solution of $(x^2-1)\,y'' + 4x\,y' + 2y = 6x$, given that $y_1=x$ and $y_2=\dfrac{x^2+x+1}{x+1}$ are two particular solutions of this non-homogeneous equation.

Step 1: Verify $y_1=x$

With $y_1'=1$, $y_1''=0$: $(x^2-1)(0)+4x(1)+2x = 6x.$ $\checkmark$

Step 2: Verify $y_2$

Polynomial division simplifies $y_2$ nicely: $x^2+x+1=(x+1)x+1$, so

\[ y_2 = x+\frac1{x+1}, \qquad y_2'=1-\frac1{(x+1)^2}, \qquad y_2''=\frac2{(x+1)^3}. \]

Substituting, and using $x^2-1=(x-1)(x+1)$: \[\frac{2(x-1)}{(x+1)^2} + 4x-\frac{4x}{(x+1)^2} + 2x+\frac2{x+1} = \frac{2(x-1)-4x}{(x+1)^2}+\frac2{x+1}+6x = -\frac{2}{x+1}+\frac2{x+1}+6x = 6x.\ \checkmark\]

Step 3: A homogeneous solution from the difference

Since $y_1$ and $y_2$ both solve the same non-homogeneous equation, their difference solves the associated homogeneous equation: \[y_h = y_2-y_1 = \left(x+\frac1{x+1}\right)-x = \frac1{x+1}.\]

Verify: with $y_h'=-\dfrac1{(x+1)^2}$, $y_h''=\dfrac2{(x+1)^3}$, \[(x^2-1)\cdot\frac2{(x+1)^3}+4x\!\left(-\frac1{(x+1)^2}\right)+\frac2{x+1} = \frac{2(x-1)-4x}{(x+1)^2}+\frac2{x+1} = -\frac2{x+1}+\frac2{x+1}=0.\ \checkmark\]

Step 4: Reduction of order on $y_h=1/(x+1)$

Set $y=y_hv$. Since $y_h$ solves the homogeneous equation, the $v$-terms cancel automatically upon substitution, leaving only the coefficients of $v'$ and $v''$: \[v'\text{-coefficient: } 2(x^2-1)\!\left(-\frac1{(x+1)^2}\right)+4x\cdot\frac1{x+1} = -\frac{2(x-1)}{x+1}+\frac{4x}{x+1} = \frac{2x+2}{x+1} = 2,\]

\[ v''\text{-coefficient: } (x^2-1)\cdot\frac1{x+1} = x-1. \] So the reduced equation is remarkably simple: $(x-1)\,v'' + 2v' = 0.$

Step 5: Solve for $v$

With $p=v'$: $\dfrac{dp}p=-\dfrac2{x-1}\,dx \implies p=\dfrac{A}{(x-1)^2}$. Integrating: $v = -\frac{A}{x-1}+B.$

Step 6: Second homogeneous solution

\[ y = \frac{v}{x+1} = -\frac{A}{(x-1)(x+1)}+\frac{B}{x+1} = -\frac{A}{x^2-1}+\frac{B}{x+1}. \]

The $B$ part reproduces $y_h$; the new solution is $\tilde y_2=\dfrac1{x^2-1}$. Verify: with $\tilde y_2'=-\dfrac{2x}{(x^2-1)^2}$, $\tilde y_2''=\dfrac{6x^2+2}{(x^2-1)^3}$,

\[ (x^2-1)\cdot\frac{6x^2+2}{(x^2-1)^3}+4x\!\left(-\frac{2x}{(x^2-1)^2}\right)+\frac2{x^2-1} = \frac{-2x^2+2}{(x^2-1)^2}+\frac2{x^2-1} = -\frac2{x^2-1}+\frac2{x^2-1}=0.\ \checkmark \]

Step 7: Homogeneous general solution

\[ Y_h = \frac{C_1}{x+1}+\frac{C_2}{x^2-1}, \qquad W\!\left(\frac1{x+1},\frac1{x^2-1}\right) = -\frac1{(x^2-1)^2}\neq0. \]

Step 8: Particular solution

Step 1 already showed that $y_1=x$ is itself a particular solution of the full non-homogeneous equation, so $y^*=x$ directly — no further work is needed.

Step 9: General solution

The solution is recorded in the result below.