In the equation (1+x^2)y'' - 2xy' + (ax^2+bx+c)y = 0…

In the equation $(1+x^2)y'' - 2xy' + (ax^2+bx+c)y = 0$, determine the constants $a,b,c$ so that $y=e^{mx}$ is a particular solution, and find the general solution.

Step 1: Forcing $y=e^{mx}$ to solve the equation

With $y=e^{mx}$, $y'=me^{mx}$, $y''=m^2e^{mx}$. Substituting and dividing through by $e^{mx}\neq0$ gives a polynomial identity in $x$: \[(1+x^2)m^2 - 2mx + ax^2+bx+c = 0 \;\Longrightarrow\; (m^2+a)x^2 + (b-2m)x + (m^2+c) = 0.\]

For this to hold for every $x$, each coefficient must vanish: \[a = -m^2, \qquad b = 2m, \qquad c = -m^2.\]

With these values the equation reads

\[ (1+x^2)y'' - 2xy' - \bigl[m^2(1+x^2) - 2mx\bigr]y = 0, \tag{H} \]

and $y_1=e^{mx}$ is one solution by construction.

Step 2: Reduction of order: eliminating the known solution

Set $y=e^{mx}v(x)$, so $y'=e^{mx}(v'+mv)$ and $y''=e^{mx}(v''+2mv'+m^2v)$. Substituting into $(H)$ and dividing by $e^{mx}$: $(1+x^2)(v''+2mv'+m^2v) - 2x(v'+mv) - \bigl[m^2(1+x^2)-2mx\bigr]v = 0$.

The $v$-terms cancel exactly — $m^2(1+x^2)v$ against $-m^2(1+x^2)v$, and $-2mxv$ against $+2mxv$ — because $y_1$ itself solves the homogeneous equation. What remains is a second-order equation with no $v$ term:

\[ (1+x^2)v'' + \bigl[2m(1+x^2) - 2x\bigr]v' = 0. \]

Step 3: Solving for $w=v'$

This is first-order and separable in $w=v'$: \[\frac{dw}{w} = \left(-2m + \frac{2x}{1+x^2}\right)dx \;\Longrightarrow\; \ln w = -2mx + \ln(1+x^2) + \mathrm{const},\]

so $w = v' = A(1+x^2)e^{-2mx}$.

Step 4: Integrating twice to recover $v$

Integrating $v'=A(1+x^2)e^{-2mx}$ by parts (the standard reduction formulas for $\int x^ke^{-2mx}dx$, $k=0,2$) gives

\[ v(x) = -\frac{A}{4m^3}\,e^{-2mx}\bigl(2m^2x^2+2mx+2m^2+1\bigr) + B. \]

This is confirmed directly: differentiating the right-hand side reproduces $A(1+x^2)e^{-2mx}$ term for term, with every $x$-independent piece cancelling — so the antiderivative is exact, without needing to display the intermediate by-parts bookkeeping.

Step 5: The second solution and the general solution

Since $y=e^{mx}v$, the term $Be^{mx}$ merely reproduces $y_1$ (up to a constant), while the other term supplies a genuinely new solution. Absorbing the constant $-A/(4m^3)$ into the arbitrary coefficient, the second solution is

\[ y_2 = \bigl(2m^2x^2+2mx+2m^2+1\bigr)e^{-mx}. \]

Hence the general solution of the original equation, with $a=-m^2$, $b=2m$, $c=-m^2$, is

\[ y = C_1e^{mx} + C_2\bigl(2m^2x^2+2mx+2m^2+1\bigr)e^{-mx}. \]

Step 6: The special case $m=0$

When $m=0$, the coefficients collapse to $a=b=c=0$ and the equation becomes $(1+x^2)y''-2xy'=0$. This is separable in $p=y'$: $dp/p = 2x/(1+x^2)\,dx$ gives $p=A(1+x^2)$, and integrating once more,

\[ y = A\left(x+\frac{x^3}{3}\right) + B, \]

i.e. $y=C_1+C_2\bigl(x+x^3/3\bigr)$, the degenerate limit of the general formula as $m\to0$.

Step 7: Verification

Both $y_1=e^{mx}$ and $y_2=(2m^2x^2+2mx+2m^2+1)e^{-mx}$ satisfy $(1+x^2)y''-2xy'+(ax^2+bx+c)y=0$ exactly with $a=-m^2,\,b=2m,\,c=-m^2$; since $y_2$ was built by an exact reduction of order — every cancellation in Step 2 is algebraic, and $v'$ was confirmed to reproduce $A(1+x^2)e^{-2mx}$ by direct differentiation in Step 4 — no residual survives. For $m>0$, $y_1=e^{mx}$ grows without bound while $y_2\sim2m^2x^2e^{-mx}$ decays to zero, so the two solutions cannot be proportional: this asymptotic mismatch confirms their linear independence.

\[a=-m^2,\ \ b=2m,\ \ c=-m^2, \qquad y = C_1e^{mx} + C_2\bigl(2m^2x^2+2mx+2m^2+1\bigr)e^{-mx}\]