Repeated Roots — Question 1

Question 1 – Solution

Strategy. Remove the common exponential in this constant-coefficient equation and determine what equation remains for its multiplier.

Step 1: Identify the repeated root and lost freedom. The characteristic polynomial is $(2r-3)^2$, so $\boxed{r=3/2}$ is repeated. The student’s family equals $(C_1+C_2)e^{3t/2}$ and has only one effective constant. It always satisfies $y'=(3/2)y$, so at $t=2$ a value of $-1$ would force slope $-3/2$, not $1/2$.

Step 2: Derive the missing multiplier. With $x=t-2$ and $y=e^{3x/2}u$, $$y'=e^{3x/2}(u'+3u/2),\qquad y''=e^{3x/2}(u''+3u'+9u/4).$$ Substitution gives $4y''-12y'+9y=4e^{3x/2}u''$. Thus $u''=0$ and $u=A+Bx$, yielding $$\boxed{y=(A+B(t-2))e^{3(t-2)/2}}.$$

Step 3: Apply and verify the data. At $t=2$, $y=A$ and $y'=B+3A/2$. Hence $A=-1$, $B=2$, so $$\boxed{y=(-1+2(t-2))e^{3(t-2)/2}}.$$ The multiplier has zero second derivative, making the original residual zero by Step 2. At 2, the value is $-1$ and the derivative is $2-3/2=1/2$.

Step 4: Establish the two independent data freedoms. For arbitrary data $y(2)=a$, $y'(2)=b$, we obtain $A=a$, $B=b-3a/2$, uniquely. The transformation is invertible because the exponential never vanishes, and integrating $u''=0$ gives all multipliers. Alternatively, continuous-coefficient IVP uniqueness proves completeness after the data map is verified.

The function $(t-2)e^{3(t-2)/2}$ has zero value but nonzero derivative at the base point, allowing slope to change without changing the prescribed value. A second copy of the same exponential cannot do that.