Basic Concepts for nth Order Linear Equations — Question 1

Question 1 – Solution

Strategy. Separate the equation’s order from the intervals on which its highest derivative can be isolated.

Step 1: Classify and normalize. This is a fourth-order, linear, nonhomogeneous equation: coefficients depend only on $x$, and the forcing $\ln x$ is not identically zero. For $x>0$, $x\ne 1$, $$\boxed{y^{(4)}+\frac{2x}{x^2-1}y'''-\frac 3{x^2-1}y' =\frac{\ln x}{x^2-1}.}$$ The coefficients of $y''$ and $y$ are zero. The logarithm excludes $x\le 0$; normalization additionally excludes $1$.

Step 2: Apply the theorem on its proper interval. All normalized coefficients and the forcing are continuous on $\boxed{(0,1)}$, the largest such interval through $1/2$. The linear IVP theorem gives one solution for arbitrary values of $y,y',y'',y'''$ at that ordinary point. These four data specify the fourth-order initial state. The theorem does not assert that this particular solution cannot extend beyond the interval; it simply gives no guarantee across $1$.

Step 3: Check the singular-point constraint. At $x=1$, any $C^4$ solution must satisfy $$\boxed{2y'''(1)-3y'(1)=0.}$$ The proposed data give $2-6=-4$, not zero. They are impossible for a classical solution of the original equation near $1$.

Step 4: Interpret compatibility without overclaiming. With $y'''(1)=3$, the same test gives $6-6=0$. This removes that contradiction, but alone proves neither existence nor uniqueness; the normalized IVP theorem still does not apply at $1$. The coefficient of $y^{(4)}$ is nonzero at nearby points other than $1$, so the equation remains fourth order on a neighborhood. At the isolated point, its value imposes a lower-derivative constraint instead of determining $y^{(4)}(1)$.