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== 1.1#h(0.6em)Solution Families


    #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
#emph[Objectives]


      - Recognize #math.equation(block: false, alt: "y prime equals k y")[$y^(′) = k y$] as the statement "rate is proportional to amount" and #math.equation(block: false, alt: "y equals a e to the power k t")[$y = a e^(k t)$] as everything that statement allows.
- Explain what each constant in a solution family controls, and how an initial condition selects one member of the family.
- Solve an initial value problem for exponential growth or decay and read off doubling times and half-lives.
]


    The shortest differential equation with something to say is


    #math.equation(block: true, alt: "the fraction d y over d t equals k y .")[$frac(d y, d t) = k y .$]


    In words: the rate of change of #math.equation(block: false, alt: "y")[$y$] is proportional to #math.equation(block: false, alt: "y")[$y$] itself. Money earning continuous interest obeys it, a bacterial culture with unlimited food obeys it, a radioactive sample obeys it with #math.equation(block: false, alt: "k")[$k$] negative. The equation does not mention any particular starting amount — and that is the point. A differential equation constrains the #emph[shape] of a solution, not its starting place, so it is satisfied not by one function but by a whole family of them:


    #math.equation(block: true, alt: "y open parenthesis t close parenthesis equals a e to the power k t ,")[$y ( t ) = a e^(k t) ,$]


    one member for every choice of the constant #math.equation(block: false, alt: "a")[$a$]. Substitute back to check: the derivative of #math.equation(block: false, alt: "a e to the power k t")[$a e^(k t)$] is #math.equation(block: false, alt: "k times a e to the power k t")[$k · a e^(k t)$], which is exactly #math.equation(block: false, alt: "k")[$k$] times the function. The constant #math.equation(block: false, alt: "a")[$a$] is not decoration; it is the answer to a question the differential equation never asked, namely #emph[where does the solution start?] Setting #math.equation(block: false, alt: "t equals 0")[$t = 0$] gives #math.equation(block: false, alt: "y open parenthesis 0 close parenthesis equals a")[$y ( 0 ) = a$].


    The figure below puts the whole family under your hands. The slider #math.equation(block: false, alt: "a")[$a$] is the initial value #math.equation(block: false, alt: "y open parenthesis 0 close parenthesis")[$y ( 0 )$] and the slider #math.equation(block: false, alt: "b")[$b$] plays the role of the rate constant #math.equation(block: false, alt: "k")[$k$].


    
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      The exponential family #math.equation(block: false, alt: "y equals a e to the power b t")[$y = a e^(b t)$]. The slider #math.equation(block: false, alt: "a")[$a$] sets the starting value #math.equation(block: false, alt: "y open parenthesis 0 close parenthesis")[$y ( 0 )$]; the slider #math.equation(block: false, alt: "b")[$b$] sets the growth rate. Every curve that these two dials can produce is a solution of #math.equation(block: false, alt: "y prime equals b y")[$y^(′) = b y$].
    
    
=== Explore


    + #strong[Observe.] Set #math.equation(block: false, alt: "b equals 0.5")[$b = 0.5$] and sweep #math.equation(block: false, alt: "a")[$a$] from #math.equation(block: false, alt: "3")[$3$] down through #math.equation(block: false, alt: "0")[$0$] to #math.equation(block: false, alt: "minus 3")[$− 3$]. Every curve in the sweep passes through the point #math.equation(block: false, alt: "open parenthesis 0 , a close parenthesis")[$( 0 , a )$]. What single point does the entire family have in common when #math.equation(block: false, alt: "a equals 0")[$a = 0$]?
+ #strong[Predict.] With #math.equation(block: false, alt: "a equals 2")[$a = 2$] fixed, what will happen to the curve as you drag #math.equation(block: false, alt: "b")[$b$] from #math.equation(block: false, alt: "0.5")[$0.5$] down through #math.equation(block: false, alt: "0")[$0$] to #math.equation(block: false, alt: "minus 1")[$− 1$]? Sketch your guess for all three regimes before touching the slider.
+ #strong[Verify.] Now drag #math.equation(block: false, alt: "b")[$b$]. At #math.equation(block: false, alt: "b equals 0")[$b = 0$] the equation reads #math.equation(block: false, alt: "y prime equals 0")[$y^(′) = 0$] — which functions satisfy that, and is the flat line you see one of them?
+ #strong[Observe.] Make the curve decay: #math.equation(block: false, alt: "a equals 3")[$a = 3$], #math.equation(block: false, alt: "b equals minus 1")[$b = − 1$]. Read off roughly where the curve has dropped to half its starting height. Then set #math.equation(block: false, alt: "b equals minus 2")[$b = − 2$] and read the half-height time again. How did doubling the decay rate change the half-life?
+ #strong[Predict, then verify.] Can you find slider settings that make the curve pass through the point #math.equation(block: false, alt: "open parenthesis 1 , 2 close parenthesis")[$( 1 , 2 )$]? There are many — find two visibly different ones, and explain why the initial value problem #math.equation(block: false, alt: "y open parenthesis 0 close parenthesis equals 2")[$y ( 0 ) = 2$] has only one.


    
=== Connect


    An #emph[initial value problem] bundles a differential equation with a starting fact: #math.equation(block: false, alt: "y prime equals k y")[$y^(′) = k y$] together with #math.equation(block: false, alt: "y open parenthesis 0 close parenthesis equals y sub 0")[$y ( 0 ) = y_(0)$]. The equation supplies the family #math.equation(block: false, alt: "y equals a e to the power k t")[$y = a e^(k t)$]; the initial condition picks the member with #math.equation(block: false, alt: "a equals y sub 0")[$a = y_(0)$]; and that member is the only one. This one-condition-one-solution pattern is the normal situation for first-order equations, and it is why the sliders in the figure feel like coordinates: the pair #math.equation(block: false, alt: "open parenthesis a , b close parenthesis")[$( a , b )$] is a complete address for a curve. Exploration step 5 shows the flip side — a point away from #math.equation(block: false, alt: "t equals 0")[$t = 0$] does not pin down a curve unless the rate is also known, which is why the problem statement must say #emph[which] equation and #emph[which] starting value.


    #examplebox("Example 1")[An initial value problem][
A culture obeys #math.equation(block: false, alt: "y prime equals 0.5 y")[$y^(′) = 0.5 y$] with #math.equation(block: false, alt: "y open parenthesis 0 close parenthesis equals 3")[$y ( 0 ) = 3$], in thousands of cells and hours. Find #math.equation(block: false, alt: "y open parenthesis t close parenthesis")[$y ( t )$] and determine when the culture reaches #math.equation(block: false, alt: "6")[$6$] thousand cells.


      The family is #math.equation(block: false, alt: "y equals a e to the power 0.5 t")[$y = a e^(0.5 t)$], and #math.equation(block: false, alt: "y open parenthesis 0 close parenthesis equals 3")[$y ( 0 ) = 3$] forces #math.equation(block: false, alt: "a equals 3")[$a = 3$], so #math.equation(block: false, alt: "y open parenthesis t close parenthesis equals 3 e to the power 0.5 t")[$y ( t ) = 3 e^(0.5 t)$]. Setting #math.equation(block: false, alt: "3 e to the power 0.5 t equals 6")[$3 e^(0.5 t) = 6$] gives #math.equation(block: false, alt: "e to the power 0.5 t equals 2")[$e^(0.5 t) = 2$], so #math.equation(block: false, alt: "0.5 t equals ln 2")[$0.5 t = ln 2$] and #math.equation(block: false, alt: "t equals 2 ln 2 approximately equals 1.39")[$t = 2 ln 2 ≈ 1.39$] hours. Check it on the figure: with #math.equation(block: false, alt: "a equals 3")[$a = 3$] and #math.equation(block: false, alt: "b equals 0.5")[$b = 0.5$], the curve crosses height #math.equation(block: false, alt: "6")[$6$] just short of #math.equation(block: false, alt: "t equals 1.4")[$t = 1.4$]. Note the doubling time did not depend on the starting amount — only on the rate.
]