One forced initial value problem solved two ways, and a first-order one solved a third. Undetermined coefficients guesses a particular solution, adds the homogeneous one and fits two constants. The Laplace transform puts the starting values in at the first line and ends with a table read backwards. The two give the same function, and the reason is that an initial value problem has one solution. Each method has a kind of problem it is best at.
Core ideas
Three ideas: each method's route, why the answers must agree, and how to choose.
Guess a particular solution y_p with the shape of the forcing, add the homogeneous solution y_h with two unknown constants, and fit the constants to the starting values with a 2×2 system. The constants are found last.
Transform both sides, with y(0) and y′(0) inside the derivative rules, and solve for Y. Partial fractions and the table finish the job. No general solution is written and no system for constants is solved.
Two solutions of the same initial value problem differ by a solution of the homogeneous equation that starts at zero, and that is zero. So the routes return the same function. The lab's residual 0 is the test on any single answer, starting values included.
The material
The constant-force equation by both second-order routes, a proof that they agree, the first-order case by a third route, and when to choose which.
Take y″ + 3y′ + 2y = 4 with y(0) = 0 and y′(0) = 0. The forcing is constant, so the first route guesses a constant. If y_p = K, then y_p′ = 0 and y_p″ = 0, and the equation reads 2K = 4, so y_p = 2. The homogeneous equation has the characteristic roots −1 and −2, so y_h = C₁·e^(−t) + C₂·e^(−2t). The general solution is y = 2 + C₁·e^(−t) + C₂·e^(−2t).
Fitting the constants takes two equations. From y(0) = 0: 2 + C₁ + C₂ = 0, so C₁ + C₂ = −2. From y′(0) = 0, with y′ = −C₁·e^(−t) − 2C₂·e^(−2t): −C₁ − 2C₂ = 0. Adding the two equations gives −C₂ = −2, so C₂ = 2 and C₁ = −4.
The second route starts at the same equation. Transforming gives (s² + 3s + 2)·Y = 4/s, so Y = 4/(s·(s + 1)·(s + 2)). The roots give the coefficients: 4/(1·2) = 2 at s = 0, 4/((−1)·1) = −4 at s = −1, and 4/((−2)·(−1)) = 2 at s = −2. So Y = 2/s − 4/(s + 1) + 2/(s + 2), and the table backwards gives y = 2 − 4·e^(−t) + 2·e^(−2t).
The same function from two directions. The lab prints this solution and its residual in the original equation and starting values, residual 0, so the check does not rest on reading two lines and judging them alike.
If y₁ and y₂ both satisfy y″ + 3y′ + 2y = 4 with y(0) = 0 and y′(0) = 0, then y₁ = y₂.
Let d = y₁ − y₂. Because the equation is linear, the 4 cancels and d satisfies the homogeneous equation, with d(0) = 0 and d′(0) = 0.
So d = C₁·e^(−t) + C₂·e^(−2t), and the starting values give C₁ + C₂ = 0 and −C₁ − 2C₂ = 0. Adding them gives −C₂ = 0, so C₂ = 0 and C₁ = 0. The other kinds of root end the same way, because the 2×2 system for the constants has a nonzero determinant, the Wronskian at 0 of “Superposition and the Wronskian” in Second-Order Linear Equations.
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The equation y′ + 2y = e^t with y(0) = 0 has a third route, the integrating factor. With μ = e^(2t) the left side times μ is (e^(2t)·y)′, and the right side is e^(3t). So e^(2t)·y = e^(3t)/3 + C, and y(0) = 0 gives C = −1/3. The result is y = e^t/3 − e^(−2t)/3.
The transform gives (s + 2)·Y = 1/(s − 1), so Y = 1/((s − 1)(s + 2)). At s = 1 the coefficient is 1/3, and at s = −2 it is −1/3. Undetermined coefficients guesses A·e^t: A + 2A = 1, so A = 1/3, and the starting value gives the constant −1/3. All three return e^t/3 − e^(−2t)/3.
For the cosine-force preset, y″ + 4y = 3cos(t) from rest, undetermined coefficients guesses A·cos(t): −A + 4A = 3, so A = 1, and the starting values give C₁ = −1 and C₂ = 0. The transform gives Y = 3s/((s² + 1)(s² + 4)) = s/(s² + 1) − s/(s² + 4). Both give y = cos(t) − cos(2t).
Interactive
Transforms, partial fractions and solutions, exactly
Method
Do both routes, compare, and let substitution decide any disagreement.
Guess y_p with the forcing's shape, solve for its coefficient, add y_h with two constants.
Put y(0) and y′(0) into the general solution and its derivative, and solve the 2×2 system.
Transform both sides with the starting values inside, solve for Y, decompose and invert.
Match the two functions term by term. A mismatch is an arithmetic slip in one route.
Put the suspect answer into the equation and the starting values. The residual must be zero.
The first preset: the constants fitted on one side and the fractions on the other.
Both lines end at the same function. Check the starting values once: y(0) = 2 − 4 + 2 = 0 and y′(0) = 4 − 4 = 0. The lab reports residual 0 for the equation.
The two routes differ in where the work falls. The first found the constants last, from a 2×2 system. The second found them as the coefficients of the fractions, and the starting values never needed a separate step.
You should be able to fit the constants from a 2×2 system, decompose the transform of the answer, compare the two results term by term, and verify the answer by substitution.
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