Basic Concepts — Question 1

Question 1 – Solution

Strategy. Expand derivatives before classifying, and distinguish an equation from its normalized form at a zero leading coefficient.

Step 1: Expand and classify. The product rule gives $(ty')'=ty''+y'$. Hence A is $$ty''+y'+y=\sin t.$$ The classification is $$\begin{array}{c|c|l} \text{equation}&\text{order}&\text{type}\\\hline \mathrm A&2&\text{linear, nonhomogeneous}\\ \mathrm B&2&\text{nonlinear}\\ \mathrm C&2&\text{nonlinear}\\ \mathrm D&2&\text{linear, homogeneous} \end{array}$$ A is a second-order equation whose leading coefficient vanishes at one point; its order as an equation is not reclassified solely at that point.

Step 2: Identify what is being squared. In D, $t^2$ is a known coefficient, while $y$ occurs to the first power. In B and C the unknown function or its derivative is squared. A zero right-hand side does not remove that nonlinearity. Under the stated convention, neither B nor C is a homogeneous linear equation.

Step 3: Normalize on a regular interval. For $t\ne 0$, division by $t$ yields $$\boxed{y''+\frac 1t y'+\frac 1t y=\frac{\sin t}{t}}.$$ All three normalized coefficients are continuous on $\boxed{(0,\infty)}$, the largest open interval containing 1 with this property. Although $\sin t/t$ has a removable singularity at zero, $1/t$ does not.

Step 4: Test the original equation at zero. For a $C^2$ solution, the undivided equation at zero requires $$0\cdot y''(0)+y'(0)+y(0)=0, \qquad \boxed{y'(0)+y(0)=0}.$$ The proposed data give $0+1\ne 0$, so no such solution exists. The condition is necessary; deriving it alone is not an existence or uniqueness proof for every compatible pair. Division by the leading coefficient must not hide this compatibility requirement.