Linear Equations — Question 2

Question 2 – Solution

Strategy. Use $x^2$ as integrating factor, then retain a definite-integral form to analyze signs on both sides of the initial point.

Step 1: Solve. Dividing by $x$ gives $y'+2y/x=x\ln x$. With $\mu=x^2$, $$(x^2y)'=x^3\ln x.$$ Integration by parts, $u=\ln x$, $dv=x^3dx$, yields $$\int x^3\ln x\,dx=\frac{x^4}{4}\ln x-\frac{x^4}{16}+C.$$ The initial value forces $C=1/16$, hence $$\boxed{y=\frac{x^2}{4}\ln x-\frac{x^2}{16}+\frac{1}{16x^2},\qquad x>0.}$$

Step 2: Verify. Substitution at $1$ gives zero. Differentiating $x^2y$ gives $x^3\ln x$, or $x^2y'+2xy=x^3\ln x$; division by $x>0$ recovers the original equation. The logarithm and coefficient division restrict the interval to $\boxed{(0,\infty)}$.

Step 3: Sign and minimum. Equivalently, $$x^2y(x)=\int_1^x t^3\ln t\,dt.$$ For $x>1$ the integral is positive. For $0<x<1$, the integrand is negative between $x$ and $1$, but the reversed limits make the integral positive. Since $x^2>0$, $y(x)>0$ for $x\ne 1$. Thus $\boxed{\min_{x>0}y=0\text{ only at }x=1}$. The sign of the forcing alone does not determine the sign of the solution.