A family of solutions has a constant in it, and a single curve needs the constant pinned. An initial value does that: it says the solution passes through a given point. A first-order equation has one constant and needs one value; a second-order equation has two and needs two. The constant is found by substituting the value, and the finished solution is checked by the same residual.
Core ideas
Three ideas: what an initial value is, how many a given equation needs, and why it selects a curve and says nothing about where the solution begins.
y(1) = 5 says that the solution takes the value 5 at t = 1. Of all the curves t² + C it leaves exactly one, the one that goes through (1, 5). The equation together with the value is an initial value problem, and it is the equation and the value that have an answer, not the equation alone.
Undoing one derivative brings one constant, which is what you met as the constant of integration. A first-order equation therefore has a family with one constant and needs one value. A second-order equation undoes two derivatives, carries two constants, and needs two: the value of y and the value of y′ at the same t.
Put the given t and y into the family and solve the resulting equation for the constant: 1² + C = 5 gives C = 4. The finished function is then run through the residual test as before. The value fixes the curve; it does not make a wrong family right.
The material
How a value selects a curve, the count of values by order, three families, and where the lab stops.
A differential equation together with values of the unknown, and for a second-order equation of its derivative, at one starting t = t₀. For first order the problem is y′ = f(t, y) with y(t₀) = y₀.
A solution of the problem is a solution of the equation that also takes the stated values at t₀.
Take y′ = 2t with y(1) = 5. The family of solutions is y = t² + C. The value says that at t = 1 the function is 5, and t² + C at t = 1 is 1 + C, so 1 + C = 5 and C = 4. The solution of the problem is y = t² + 4. The lab reports this as C = 4 from y(1) = 5 and the residual stays 0: fixing a constant does not leave the family of solutions, it picks a member.
The same procedure works for any family linear in its constants. For y′ = 3y the family is C·e^(3t), and y(0) = 2 gives C·e⁰ = 2, so C = 2 and the solution is 2e^(3t). The value is taken at t = 0 because that is where the lab can evaluate an exponential exactly, since e⁰ = 1: at any other t the value would need an irrational number, and the lab says that it is refusing to fit there.
A second-order equation has two constants, and two values give two equations for them. Take y″ − y′ − 2y = 0, whose two exponentials “Checking a Proposed Solution” verified: the family is C·e^(2t) + D·e^(−t), with derivative 2C·e^(2t) − D·e^(−t). At t = 0 both exponentials are 1, so y(0) = C + D and y′(0) = 2C − D. The values y(0) = 1 and y′(0) = −2 give C + D = 1 and 2C − D = −2; adding them, 3C = −1, so C = −1/3 and D = 4/3, and the lab prints C = −1/3, D = 4/3. Two values, two equations, one small system.
For y″ + y = 0 the family is C·cos(t) + D·sin(t), with derivative −C·sin(t) + D·cos(t). That sine and cosine are each other's rates, with the one sign change, is taken on trust here and demonstrated in “Sine, Cosine and Their Rates” in Second-Order Linear Equations; what this fit needs is only their values at t = 0, where the cosine is 1 and the sine is 0, so y(0) = C and y′(0) = D. The values y(0) = 1 and y′(0) = −2 therefore give C = 1 and D = −2 at once, and the solution is cos(t) − 2·sin(t). One value would leave the other constant free.
Give the lab y″ + y = 0 with only y(0) = 1. It declines to fit anything: the constants tile shows a dash, and the banner says that two constants need two values, y(t₀) and y′(t₀).
By hand the situation is clear: y(0) = 1 forces C = 1 and says nothing about D. Every function cos(t) + D·sin(t) is a solution that passes through (0, 1), and they differ in how steeply they pass. The second value, the rate at the start, is the missing piece of information, and supplying it is what selects one.
It is tempting to read y(1) = 5 as saying that the solution begins at t = 1. It does not. The curve t² + 4 is defined on both sides of t = 1, and what the value does is identify which curve. The point t₀ is simply where the information is given; the solution runs both ways from it.
One fact about initial value problems is used on trust at this stage. For the equations in this course, a starting point selects exactly one solution. “Blow-Up and the Interval of Existence” states the condition under which that is true and shows an equation where the solution stops existing; the lab, here, only fits the constants.
Interactive
Substitute, simplify, compare with zero
Method
Four moves, once the family of solutions is known.
One constant for a first-order equation, two for a second-order one. If you cannot name the constants, you do not yet have a general solution.
Put t₀ and y₀ into the family, and for a second-order equation put t₀ and y′(t₀) into its derivative. Each value gives one equation for the constants.
One equation for one constant is arithmetic. Two equations for two constants are a small linear system: C + D = 1 and 2C − D = −2 for the exponential pair, solved by adding them, as in Algebra's Systems and Matrices. For the sine and cosine pair the system arrives already solved, because each value at t = 0 names one constant on its own.
Write the solution with the constants in place, run the residual test on it, and confirm that it takes the stated values. Both must hold.
The family is y = t² + C; find the constant and check the answer.
Two checks, and both are needed. The residual says the function solves the equation, and the value says it is the member that goes through the stated point. A function can pass either test and fail the other: t² solves the equation and misses the point, and 5 hits the point and solves nothing.
For a rehearsal, take the same equation with y(2) = 1. The substitution is 4 + C = 1, so C = −3, and the lab will say so.
You should be able to write the family of solutions with its constants, substitute a starting value to solve for the constant of a first-order equation, solve the two equations for the constants of a second-order one, state that the numbers needed are one and two, and run the residual test on the result.
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