Reduction of Order — Question 1

Question 1 – Solution

Strategy. Factor out the known solution, then normalize the new solution by its initial slope.

Step 1: Reduce the order. Product differentiation gives $$y'=y_1'v+y_1v',\qquad y''=y_1''v+2y_1'v'+y_1v''.$$ The coefficient of $v$ is zero because $y_1$ solves the equation. Hence $$y_1v''+(2y_1'+py_1)v'=0,\qquad w'+(2y_1'/y_1+p)w=0.$$

Step 2: Integrate twice. An integrating factor is $y_1^2e^P$, so $$w=C\frac{e^{-P}}{y_1^2},\qquad y=y_1(t)[A+C\int_{t_0}^t\frac{e^{-P(s)}}{y_1(s)^2}\,ds].$$ Both integrations contribute a constant. The formula includes $w=0$.

Step 3: Normalize the companion. Set $$\boxed{z(t)=y_1(t_0)y_1(t)\int_{t_0}^t\frac{e^{-P(s)}}{y_1(s)^2}\,ds.}$$ The integral vanishes at $t_0$. Since $P(t_0)=0$, differentiation gives $z'(t_0)=1$. If $z=ky_1$, its zero value forces $k=0$, contradicting this slope.

Step 4: Match arbitrary data. The two constants are fixed by $$\boxed{y=\frac{a}{y_1(t_0)}y_1+ [b-\frac{a y_1'(t_0)}{y_1(t_0)}]z.}$$ This is the unique solution on $I$ by the regular linear initial-value theorem. Division by $y_1$ and the integral kernel require $y_1\ne 0$. A zero of the seed may obstruct this substitution without obstructing the original equation.