Lesson · Differential Equations and Euler's Method · What a differential equation is

Checking a Proposed Solution

The residual test is the same on every equation, and it gets more work to do as equations get harder. A second-order equation needs two derivatives of the candidate; an equation with a coefficient that depends on t needs that coefficient carried through; and a nonlinear equation turns out to behave differently from the others. The test decides the first two exactly, and the equation's form tells you whether a multiple of a solution is still a solution.

Open the interactive lesson Practice

y″ − y′ − 2y = 0 with y = e^(2t) 4e^(2t) − 2e^(2t) − 2e^(2t) = 0 y = eᵗ leaves residual −2·eᵗ, never 0 agreeing at one t proves nothing linear: y, y′, y″ to the first power only

The residual on a harder equation, and linear against not

Core ideas

What this lesson establishes

Three ideas: carrying the substitution through more derivatives, why one matching point is not a check, and the one property that sorts equations.

Each derivative the equation mentions has to be computed

For y″ − y′ − 2y = 0 the candidate y = e^(2t) needs y′ = 2e^(2t) and y″ = 4e^(2t), because the derivative of e^(kt) is k·e^(kt) and a second derivative applies that twice. Put all three in and the residual is a sum of multiples of e^(2t) that either cancels or leaves a multiple of it behind.

A residual that is zero at one value of t does not make a solution

A function can satisfy an equation at a single t or at a few and still fail everywhere else. Take t·y′ = 2y and the candidate y = 5t: the residual is −5t, which is zero at t = 0 and nowhere else. The only verdict that counts is the simplified residual, and it is zero for every t or it is not.

An equation is linear when the unknown and its derivatives appear only to the first power

In a linear equation y, y′ and y″ are each multiplied by something that depends on t alone and added up, with at most a term in t alone besides. t·y′ = 2y is linear; y′ = y² is not, because y is squared. The distinction matters at once: a constant multiple of a solution of a linear equation with zero on the right is another solution, and for a nonlinear equation it usually is not.

The material

The same test on harder equations

A second-order equation, a coefficient that depends on t, a nonlinear equation, and the one place the nonlinear case differs.

Start with y″ − y′ − 2y = 0. It is second order, so a candidate needs two derivatives, and it is linear, so each of y″, y′ and y appears once with a number in front. Try y = e^(2t). Then y′ = 2e^(2t) and y″ = 4e^(2t), and the residual is 4e^(2t) − 2e^(2t) − 2e^(2t), which is (4 − 2 − 2)·e^(2t) = 0. A solution.

Now try y = eᵗ, the same shape with a different rate. Here y′ = eᵗ and y″ = eᵗ, and the residual is eᵗ − eᵗ − 2eᵗ = −2eᵗ. The coefficient −2 is not zero and eᵗ is never zero, so the residual is not zero at any t, and the candidate fails everywhere. The two exponentials differ only in the number in the exponent, and the equation accepts one and rejects the other.

y″ − y′ − 2y = 0 candidate y′ y″ residual verdict e^(2t) 2e^(2t) 4e^(2t) 0 a solution eᵗ eᵗ eᵗ −2·eᵗ not a solution

A coefficient that depends on t

The equation t·y′ = 2y has t multiplying the derivative, and the substitution carries it along. For y = 5t² the derivative is 10t, so t·y′ = 10t², and 2y = 10t², and the residual is 0. The candidate 5t² is a solution, and so would be t² or −3t²: each is a constant multiple of t², and in a linear equation with zero on the right that is allowed.

A candidate y = 5t makes a good contrast. Its derivative is 5, the left side is 5t, the right side is 10t, and the residual is −5t. At t = 0 both sides are zero, so a check at that one point would pass it. Type 5t into the lab and read the residual: it is not zero.

A nonlinear equation

Now y′ = y². Take y = 1/(1 − t). By the chain rule its derivative is 1/(1 − t)², and y² is 1/(1 − t)² as well, so the residual is 0. The lab decides this one exactly, because a ratio of polynomials in t can be differentiated without rounding, and it labels the equation nonlinear because y is squared.

Theorem · What linear buys, stated and used

In a linear equation whose right side is 0, a constant multiple of a solution is a solution, and the sum of two solutions is a solution. For a nonlinear equation neither is guaranteed.

The linear case follows from the residual being built from sums and constant multiples; the lab demonstrates it on the presets and does not prove it. The nonlinear case is easy to watch fail: take y′ = y² and double the solution to 2/(1 − t). Its derivative is 2/(1 − t)² and its square is 4/(1 − t)², so the residual is −2/(1 − t)², which is never zero (the lab prints it expanded, as −2/(t² − 2t + 1)). The doubling that was harmless in the linear equation breaks this one.

One limit of the lab belongs here. A ratio of polynomials in t, and exponential or sine and cosine candidates in a linear equation, are decided exactly. An exponential candidate in a nonlinear equation is not: the lab then evaluates the residual at five values of t in floating point and says so in the verdict, which reads that it checked at five points only. That is a spot check, and it is labelled as one.

Interactive

Checking a solution

Substitute, simplify, compare with zero

The residual on four equations

Each preset names an equation and a candidate; the order, linear and residual tiles are computed from them. After the presets, keep the equation and change the candidate: e^(-t) in the second-order equation, 5t in t y′ = 2y, and 2/(1 - t) in y′ = y^2, and read each residual.

Equation and candidate

Equation, with y' and y'' for the derivatives

Candidate y(t)

Initial values: t0 y0, or t0 y0 v0 (may be empty)

Order—

Linear?—

Residual—

Verdict—

Initial values—

Constants—

Candidates: a ratio of polynomials in t, such as 1/(1 - t), or sums of terms such as 3t^2, 2e^(-t), t cos(2t) and C e^(3t), with the constants C and D.

Method

Checking a candidate on a harder equation

The four moves of “What a Differential Equation Is”, with the extra care a harder equation needs.

Write down every derivative the equation uses

For a second-order equation that is y′ and y″ as well as y, each computed from the candidate before anything is substituted. Skipping this line is how the powers of two go missing from e^(2t).

Substitute and keep the coefficients

Put the candidate and its derivatives where y, y′ and y″ stand, including any factor of t in front of them, and leave the equation moved to one side so the residual is the whole left side.

Collect like terms and read the residual

Group the terms that share a factor, such as every multiple of e^(2t). If the coefficients sum to zero the residual is zero. If anything survives, the candidate fails, and the survivor is the residual.

Classify the equation

Look at y, y′ and y″. If each appears only to the first power, multiplied by something in t alone, the equation is linear. Anything such as y² or y·y′ makes it nonlinear.

Residuals, one-point checks and linearity

Score 0 / 0

Choose the answer you can justify.

Two exponentials in one second-order equation

Test e^(2t) and eᵗ against y″ − y′ − 2y = 0.

y = e^(2t): y′ = 2·e^(2t) y″ = 4·e^(2t) residual = 4·e^(2t) − 2·e^(2t) − 2·e^(2t) = 0 a solution y = eᵗ: y′ = eᵗ, y″ = eᵗ residual = eᵗ − eᵗ − 2·eᵗ = −2·eᵗ never zero, so not a solution

The only difference between the two is the number in the exponent, and the equation singles out 2. In Second-Order Linear Equations this becomes a method, because the right numbers are the roots of a quadratic. For now the point is that the check is mechanical and decisive.

For a rehearsal, try e^(−t). Its derivatives are −e^(−t) and e^(−t), and the residual is e^(−t) + e^(−t) − 2e^(−t), which is zero. This equation has two exponentials that work, and the lab confirms it.

Common mistakes

Accepting a function because it satisfies the equation at one value of tThe candidate 5t in t·y′ = 2y has both sides zero at t = 0, and its residual is −5t, which is zero nowhere else. A pass at a point says nothing; the whole simplified residual has to be zero.

Forgetting a derivative or its coefficient in a second-order checkFor e^(2t) the second derivative is 4e^(2t), not 2e^(2t), and the residual 4 − 2 − 2 is zero only with the 4. Write y′ and y″ as their own lines before substituting.

Assuming a multiple of a solution is always a solutionThat holds for a linear equation with zero on the right, which y″ − y′ − 2y = 0 and t·y′ = 2y are. It fails for y′ = y²: the double of 1/(1 − t) has residual −2/(1 − t)². Classify the equation first.

Check your understanding

✓

Finish when you can substitute an exponential or a power into an equation and decide it.

You should be able to compute the two derivatives of e^(kt) or a·tⁿ, substitute into a second-order or variable-coefficient equation, simplify the residual by collecting like terms, state the verdict, and say whether the equation is linear and what that allows.

Note: A residual of zero says a function is one solution, and the previous lesson showed there are usually many. The next one picks a single curve from the family by fixing the constants from starting values, and asks how many such values an equation needs. “Initial Value Problems” is where.