Find the general solution of (1-2x^2) y'' + 2y' + 4y = 0…

Find the general solution of $(1-2x^2)\,y'' + 2y' + 4y = 0$, given that a particular solution is a polynomial.

Step 1: Find the degree of the polynomial solution

Try $y_1 = x^n + \cdots$ and match the leading powers of $x$ after substitution. Since $y_1'' \sim n(n-1)x^{n-2}$, the term $-2x^2y_1''$ contributes $-2n(n-1)x^n$ at degree $n$, while $4y_1$ contributes $4x^n$ (the lower-degree pieces of $y_1''$ and $y_1'$ never reach degree $n$). Cancellation of the top coefficient requires

\[ -2n(n-1) + 4 = 0 \implies n(n-1) = 2 \implies n^2-n-2=0 \implies (n-2)(n+1)=0. \]

The polynomial (non-negative integer) root is $n=2$.

Step 2: Solve for the coefficients

Try $y_1 = x^2+bx+c$, so $y_1'=2x+b$, $y_1''=2$. Substituting: $(1-2x^2)(2) + 2(2x+b) + 4(x^2+bx+c) = 0$, $2 - 4x^2 + 4x + 2b + 4x^2+4bx+4c = 0$.

The $x^2$ terms cancel automatically; matching $x^1$ gives $4+4b=0\Rightarrow b=-1$, and matching $x^0$ gives $2+2b+4c=0 \Rightarrow 2-2+4c=0 \Rightarrow c=0$. Hence

\[ y_1 = x^2 - x. \]

Step 3: Verify $y_1$

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

Step 4: Substitute $y=y_1v$ and reduce the order

Set $y=(x^2-x)v(x)$. Then

\[ y' = (2x-1)v + (x^2-x)v', \qquad y'' = 2v + 2(2x-1)v' + (x^2-x)v''. \]

Substituting into $(1-2x^2)y''+2y'+4y=0$, the coefficient of $v$ alone reproduces the Step 3 identity and vanishes. The coefficient of $v'$ is

\[ 2(1-2x^2)(2x-1) + 2(x^2-x) = (-8x^3+4x^2+4x-2) + (2x^2-2x) = -8x^3+6x^2+2x-2, \]

and the coefficient of $v''$ is $(1-2x^2)(x^2-x)$. The reduced equation is

\[ (1-2x^2)(x^2-x)\,v'' - 2(4x^3-3x^2-x+1)\,v' = 0. \]

Step 5: Solve the first-order equation for $w=v'$

With $w=v'$, the equation is separable: $\frac{dw}{w} = \frac{2(4x^3-3x^2-x+1)}{x(x-1)(1-2x^2)}\,dx.$

Rather than clearing denominators blindly, split off the logarithmic derivative of $y_1=x(x-1)$: since $\dfrac{y_1'}{y_1}=\dfrac{2x-1}{x(x-1)}=\dfrac1x+\dfrac1{x-1}$, the right side decomposes as

\[ \frac{2(4x^3-3x^2-x+1)}{x(x-1)(1-2x^2)} = -\frac{2y_1'}{y_1} - \frac{2}{1-2x^2} = -\frac2x-\frac2{x-1}-\frac2{1-2x^2}. \]

This is confirmed by clearing denominators: writing the left side as $\dfrac{A}{x}+\dfrac{B}{x-1}+\dfrac{Cx+D}{1-2x^2}$ and matching coefficients in $2(4x^3-3x^2-x+1)=A(x-1)(1-2x^2)+Bx(1-2x^2)+(Cx+D)x(x-1)$ gives $A=-2$, $B=-2$, $C=0$, $D=-2$ — a constant numerator over $1-2x^2$, not a multiple of $x$. Integrating, and using $\displaystyle\int\frac{2\,dx}{1-2x^2}=\sqrt2\,\operatorname{artanh}(\sqrt2\,x)$ for $|x|<\tfrac1{\sqrt2}$:

\[ \ln|w| = -2\ln|x| - 2\ln|x-1| - \sqrt2\,\operatorname{artanh}(\sqrt2\,x) + \mathrm{const}, \]

\[ w = \frac{A\,e^{-\sqrt2\,\operatorname{artanh}(\sqrt2\,x)}}{x^2(x-1)^2}. \]

Step 6: Integrate for $v$

Then $v' = \frac{A\,e^{-\sqrt2\,\operatorname{artanh}(\sqrt2\,x)}}{x^2(x-1)^2}.$

This integrand has no elementary antiderivative, so $v$ is left as a quadrature: $v = A\int \frac{e^{-\sqrt2\,\operatorname{artanh}(\sqrt2\,x)}}{x^2(x-1)^2}\,dx + B.$

Step 7: General solution

Since $y=(x^2-x)v$, the general solution is

Step 8: Linear independence

By construction of reduction of order, the Wronskian obeys $W(y_1,y_2)=y_1^2w$ (Abel's identity): \[W(y_1,y_2) = (x^2-x)^2\cdot\frac{A\,e^{-\sqrt2\,\operatorname{artanh}(\sqrt2\,x)}}{x^2(x-1)^2} = A\,e^{-\sqrt2\,\operatorname{artanh}(\sqrt2\,x)} \neq 0,\] so $y_1,y_2$ form a fundamental system.