Find the general solution of xy'' - 4y' - xy = 0 , using the…

Find the general solution of $xy'' - 4y' - xy = 0$, using the substitution $y = x^2z(x)$.

Step 1: Motivation for the substitution

Repeatedly differentiating the equation $xy''-4y'-xy=0$ reveals that it reproduces its own structure with a shifted coefficient pattern after each differentiation — a self-similarity characteristic of equations that reduce to a simpler form under a power-of-$x$ substitution. This motivates trying $y=x^2z(x)$, which is confirmed below to work cleanly.

Step 2: Substituting $y=x^2z$

With $y=x^2z$, $y' = 2xz + x^2z', \qquad y'' = 2z + 4xz' + x^2z''$.

Substituting into $xy''-4y'-xy=0$: $x\bigl(2z+4xz'+x^2z''\bigr) - 4\bigl(2xz+x^2z'\bigr) - x\cdot x^2z = 0$.

\[ 2xz + 4x^2z' + x^3z'' - 8xz - 4x^2z' - x^3z = 0. \]

The $4x^2z'$ terms cancel, and $2xz-8xz=-6xz$, leaving

\[ x^3z'' - x^3z - 6xz = 0. \]

Dividing by $x$: $x^2z'' - (x^2+6)z = 0, \qquad \text{i.e.} \qquad z'' - \left(1+\frac{6}{x^2}\right)z = 0$.

Step 3: Recognizing the reduced equation

This is a modified spherical Bessel-type equation of order $n=2$ (note $n(n+1)=2\cdot3=6$), which admits elementary closed-form solutions built from $e^{\pm x}$ and inverse powers of $x$:

\[ z_1 = \left(1-\frac{3}{x}+\frac{3}{x^2}\right)e^{x}, \qquad z_2 = \left(1+\frac{3}{x}+\frac{3}{x^2}\right)e^{-x}. \]

(Direct substitution of $z_1,z_2$ into $z''-(1+6/x^2)z=0$ confirms both satisfy the equation exactly, with residual $0$ in each case.)

Step 4: Back-substitution: $y=x^2z$

Multiplying through by $x^2$ clears the fractions entirely: \[y_1 = x^2z_1 = (x^2-3x+3)e^{x}, \qquad y_2 = x^2z_2 = (x^2+3x+3)e^{-x}.\]

Step 5: Verification directly in the original equation

Rather than re-checking the intermediate $z$-equation, it is cleanest to substitute $y_1,y_2$ directly back into $xy''-4y'-xy=0$. Differentiating $y_1=(x^2-3x+3)e^x$: $y_1' = (2x-3)e^x + (x^2-3x+3)e^x = (x^2-x)e^x$.

\[ y_1'' = (2x-1)e^x + (x^2-x)e^x = (x^2+x-1)e^x. \]

Substituting into the original equation: $xy_1'' - 4y_1' - xy_1 = x(x^2+x-1)e^x - 4(x^2-x)e^x - x(x^2-3x+3)e^x$.

Collecting the polynomial coefficient of $e^x$: $(x^3+x^2-x) - (4x^2-4x) - (x^3-3x^2+3x) = x^3+x^2-x-4x^2+4x-x^3+3x^2-3x = (x^2-4x^2+3x^2) + (-x+4x-3x) = 0$. So $xy_1''-4y_1'-xy_1=0$ identically. $\checkmark$

Similarly for $y_2=(x^2+3x+3)e^{-x}$, carefully recomputing the derivative rather than assuming symmetry with $y_1$: $y_2' = (2x+3)e^{-x} - (x^2+3x+3)e^{-x} = -(x^2+x)e^{-x}$.

\[ y_2'' = -(2x+1)e^{-x} + (x^2+x)e^{-x} = (x^2-x-1)e^{-x}. \]

Substituting: $xy_2''-4y_2'-xy_2 = x(x^2-x-1)e^{-x} + 4(x^2+x)e^{-x} - x(x^2+3x+3)e^{-x}$. The polynomial coefficient of $e^{-x}$ is $(x^3-x^2-x)+(4x^2+4x)-(x^3+3x^2+3x) = (x^3-x^3)+(-x^2+4x^2-3x^2)+(-x+4x-3x)=0$. So $xy_2''-4y_2'-xy_2=0$ identically as well. $\checkmark$

Step 6: Wronskian and independence

By Abel's formula, since the equation in standard form $y''-\dfrac{4}{x}y'-y=0$ has $P(x)=-4/x$, \[W(x) = C\exp\!\left(-\int P(x)\,dx\right) = C\exp\!\left(\int \frac{4}{x}\,dx\right) = Cx^4.\]

Using the derivatives found above, $y_1'=(x^2-x)e^x$ and $y_2'=-(x^2+x)e^{-x}$, evaluate at the convenient point $x=1$: $y_1(1)=e$, $y_2(1)=7e^{-1}$, $y_1'(1)=0$, $y_2'(1)=-2e^{-1}$, so

\[ W(1) = y_1(1)y_2'(1) - y_1'(1)y_2(1) = e\cdot(-2e^{-1}) - 0\cdot 7e^{-1} = -2. \]

Since $W(1)=-2\ne0$ and $W(x)=Cx^4$ with $C=W(1)/1^4=-2$, the Wronskian $W(x)=-2x^4$ is nonzero for all $x\ne0$, confirming that $y_1,y_2$ are independent solutions throughout the domain.