Reduction of Order — Question 10

Question 10 – Solution

Strategy. The known homogeneous solution still cancels the term containing $v$, but the nonzero right-hand side remains.

Step 1: Verify the seed’s role. For $y_1=t$, the left-hand side is $0-2/t+2/t=0$. This satisfies the homogeneous equation, but cannot equal the positive forcing $t$ on the specified domain.

Step 2: Reduce with the forcing present. With $y=tv$, $$y'=v+tv',\qquad y''=2v'+tv''.$$ Substitution cancels the terms in $v$ and $v'$: $$2v'+tv''-\frac 2t(v+tv')+\frac 2t v=tv''=t.$$ Since $t>0$, $v''=1$. Equivalently the first-order reduced equation is $w'=1$, with $w=v'$. Integrating twice gives $$v=\tfrac 12t^2+At+B,\qquad \boxed{y=\tfrac 12t^3+At^2+Bt.}$$

Step 3: Impose the data. At $1$ the equations are $1/2+A+B=0$ and $3/2+2A+B=0$. Thus $A=-1$, $B=1/2$, yielding $$\boxed{y=\tfrac 12t(t-1)^2.}$$

Step 4: Verify and diagnose. In expanded form the solution is $t^3/2-t^2+t/2$, with $y'=3t^2/2-2t+1/2$ and $y''=3t-2$. Its left-hand side is $$(3t-2)+(-3t+4-1/t)+(t-2+1/t)=t.$$ At $1$ its value and slope are zero. Discarding the forcing would instead give $w'=0$, producing only $At^2+Bt$, whose left-hand side is zero. Those two terms provide the homogeneous freedom, but the cubic term is essential to match the forcing.