Linear Homogeneous Differential Equations — Question 9

Question 9 – Solution

Strategy. Solve the factors in their given order and use the product rule when reversing them.

Step 1: Reduce the original equation. Expansion gives $y'''-xy''=0$. Thus $z'-xz=0$ and $z=Ce^{x^2/2}$. Integrating twice from zero yields $$\boxed{y=A+Bx+CF(x),\qquad F(x)=\int_0^x(x-t)e^{t^2/2}\,dt.}$$ This formula holds for negative $x$ as well, with the usual oriented integral.

Step 2: Verify the fundamental set. Differentiating the integral gives $F'=\int_0^x e^{t^2/2}\,dt$ and $F''=e^{x^2/2}$, so $F'''=xF''$. The derivative matrix is triangular: $$W(1,x,F)=\det\begin{pmatrix}1&x&F\\0&1&F'\\0&0&e^{x^2/2}\end{pmatrix} =\boxed{e^{x^2/2}>0}.$$ These are three independent solutions of a regular third-order linear equation, hence a fundamental set on $\mathbb R$.

Step 3: Fit and analyze the initial data. Since $F(0)=F'(0)=0$ and $F''(0)=1$, the IVP gives $\boxed{y=F}$. Its second derivative is even, and its value and slope at zero vanish; integrating shows that $F$ is even. For $x>0$, $$F(x)-\frac{x^2}{2}=\int_0^x(x-t)(e^{t^2/2}-1)\,dt>0.$$ The inequality follows because the integrand is positive for $0<t<x$. Evenness proves the same strict inequality for $x<0$; equality holds at zero.

Step 4: Respect the order of the factors. The product rule gives $$D^2(y'-xy)=y'''-xy''-2y'.$$ Subtracting the two equations forces $y'=0$ for any common solution. Conversely, every constant solves both, so their common space consists exactly of constants. The variable coefficient creates the extra $-2y'$ term; treating $x$ as a fixed characteristic root would discard it and give an invalid method.