Reduction of Order — Question 5

Question 5 – Solution

Strategy. Compare the actual functions produced by changes in normalization, rather than counting written constants.

Step 1: Identify the kernel. The seed residual is $-2t+2t=0$. After division by $t^2$, $p=-2/t$ and $$P(t)=\int_1^t -2/s\,ds=-2\ln t,\qquad e^{-P(t)}=t^2.$$ Reduction gives $v'=C t^2/y_1^2$, which yields the specified convention.

Step 2: Compute three companions. For $y_1=t$, the integrand equals $1$, so $$Y_1=t(t-1)=t^2-t,\qquad Y_2=t(t-2)=t^2-2t.$$ For seed $3t$, the integrand is $1/9$, hence $$\widetilde Y_1=3t(t-1)/9=(t^2-t)/3.$$

Step 3: Compare freedoms. We have $Y_2=Y_1-t$ and $\widetilde Y_1=Y_1/3$. Adding a seed multiple or multiplying a companion by a nonzero constant leaves the family unchanged: every choice spans $\{At+Bt^2:A,B\in\mathbb R\}$. An additive constant inside the antiderivative produces only another multiple of the seed, already present in the family.

Step 4: Fit and verify. Since $Y_1(1)=0$, $Y_1'(1)=1$, the data give $A=2$ and $A+B=1$. Therefore $$\boxed{y=2t-(t^2-t)=3t-t^2.}$$ Its derivatives are $3-2t$ and $-2$, and $$t^2(-2)-2t(3-2t)+2(3t-t^2)=0.$$ At $t=1$ the value is $2$ and the slope is $1$. Two effective constants are sufficient and necessary for arbitrary regular initial data.