What a Differential Equation Says and What a Solution Is
A differential equation involves an unknown function and its derivatives. Check solutions on an interval. General solutions may contain arbitrary constants; initial data can determine a particular solution.
What a learner can do afterwards
- Verify that a proposed function satisfies a given equation
- State an equation's order and count independent arbitrary constants in the regular linear solution families studied
- Use an initial condition to fix the constants in a general solution
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A differential equation involves an unknown function and one or more of its derivatives. A solution has the required derivatives and satisfies the equation throughout an interval. Check it by differentiating and substituting. The derivative of e^(kx) is k e^(kx). For example, y=2e^(3x) gives y'=6e^(3x)=3y, so it solves y'=3y for every real x.
Order is the highest derivative present. The symbols y', y'' and y''' mean first, second and third derivatives. A power such as (y')² still uses only the first derivative. In a linear equation, y and its derivatives are not multiplied together or raised to powers. With continuous coefficients and forcing term, and a highest-derivative coefficient that stays nonzero, an order-n general solution has n independent constants on the interval. Call this a regular linear equation.
Initial conditions specify values at a chosen input. Substitute them into the solution family to determine its constants. In y=C e^x, y(0)=7 gives C=7 because e^0=1. For y=A cos x+B sin x, y'=-A sin x+B cos x. Since cos 0=1 and sin 0=0, y(0)=6 gives A=6, while y'(0)=-2 gives B=-2.
Integrate y'''=6: y''=6x+K1, y'=3x²+K1x+K2, and y=x³+(K1/2)x²+K2x+K3. Each integration introduces an independent constant. For y''=8, y=4x²+Ax+B; 4 is fixed, while A and B vary. Differentiate twice to get y''=8. For y''=0, y=Px+Q. A condition such as y(0)=9 fixes Q but leaves P free.
A residual is the equation's left side minus its right side after substitution. For y'=y, it is y'-y. A nonzero residual at one point disproves a proposed solution on any interval containing that point. A zero residual at one point is not enough: verification requires zero throughout the interval. To check an initial-value problem, also test each initial condition separately.
Check derivatives and initial data separately. Order is the highest derivative present. Regular linear order-n families have n independent constants. One initial value may leave constants free.
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Checking both parts of an initial-value problem
An initial-value problem asks a function to meet two kinds of requirement: a differential equation and specified starting data.
For example, y=x³-1 has derivative 3x². It therefore satisfies y'=3x² at every real input. At x=1, its value is 1³-1=0, so it also meets y(1)=0.
Changing the required value to y(1)=4 changes the problem. The same function still satisfies the differential equation, but it no longer meets the initial condition.
Check these requirements separately. First calculate the derivatives and substitute them into the equation. Then evaluate each stated initial value. A proposed solution to the whole problem must pass every check. Matching the differential equation at one selected input does not establish that it holds throughout an interval.
Which parts of a solution can vary?
Integrating y'=4x gives y=2x²+C. The coefficient 2 is fixed by the derivative. The constant C can be any real number.
If y(1)=5 is also required, substitution gives 5=2+C, so C=3. The selected function is y=2x²+3.
For y''=0, the family is y=Ax+B. A condition at x=0 on the function fixes B; a condition there on its first derivative fixes A. These data control different parts of the formula.
For a linear equation of order n, assume that every coefficient and the forcing term is continuous on an interval and that the highest-derivative coefficient never vanishes there. Its general solution has n independent arbitrary constants. Specifying the function and its first n-1 derivatives at one input in that interval determines a unique solution there. The same conclusion does not hold without further checks for arbitrary nonlinear equations.
State the interval
Consider y=1/(x+2). Its derivative is -1/(x+2)², which equals -y². Thus the formula satisfies y'=-y² wherever x is not -2.
The excluded input matters. The function is a solution on x>-2 and on x<-2, but it is not a solution on an interval containing -2. Its value and derivative do not exist at that input.
If the initial input is 0, choose a solution interval containing 0 that avoids -2. A single formula need not describe a solution across every real input.
To solve an order-n equation, a function needs the derivatives through order n throughout its solution interval. It does not need derivatives of every possible order just to satisfy that equation.
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