Verify the solution of the differential equation is a fundamental skill that every student of calculus, physics, engineering, or applied mathematics must master. Before a proposed function can be trusted as a genuine answer, it must be substituted back into the original equation and shown to satisfy it identically. This process not only confirms correctness but also deepens understanding of how differential equations model real‑world phenomena.
Introduction
Differential equations describe how quantities change with respect to one another. Solving them yields a family of functions that satisfy the relationship between a variable and its derivatives. Even so, obtaining a candidate solution—whether by separation of variables, integrating factors, characteristic equations, or numerical methods—is only half the task. The next, equally important step is to verify the solution of the differential equation by direct substitution. Verification acts as a safety net, catching algebraic slips, missed constants, or domain restrictions that might otherwise go unnoticed It's one of those things that adds up..
Understanding Differential Equations
A differential equation links an unknown function (y(x)) (or (y(t)), etc.On the flip side, ) with its derivatives. The order of the equation is the highest derivative present And that's really what it comes down to..
- First‑order: (\displaystyle \frac{dy}{dx} + P(x)y = Q(x))
- Second‑order: (\displaystyle y'' + p(x)y' + q(x)y = r(x))
Solutions can be general (containing arbitrary constants) or particular (constants fixed by initial/boundary conditions). Regardless of type, any proposed solution must satisfy the equation for all (x) in its domain The details matter here..
What Does It Mean to Verify a Solution?
To verify a solution means to plug the candidate function (and its required derivatives) into the left‑hand side of the differential equation and simplify. If the result equals the right‑hand side (often zero for homogeneous equations), the solution is valid. Symbolically, for an equation
[ F\bigl(x, y, y', y'',\dots\bigr)=0, ]
we compute
[ F\bigl(x, y_{\text{cand}}, y'{\text{cand}}, y''{\text{cand}},\dots\bigr) ]
and check whether the expression reduces to zero (or to the prescribed non‑homogeneous term) And that's really what it comes down to. That's the whole idea..
Verification is distinct from solving; it does not require finding the solution from scratch, only confirming that a given function works.
Step‑by‑Step Procedure
- Write down the candidate solution (y_c(x)) (including any constants).
- Compute the necessary derivatives (y_c', y_c'',\dots) up to the order of the equation.
- Substitute (y_c) and its derivatives into the differential equation.
- Simplify the resulting expression algebraically.
- Compare the simplified left‑hand side with the right‑hand side.
- If they match identically (for all (x) in the domain), the solution is verified.
- If not, re‑examine algebra, derivative calculations, or the original candidate.
Throughout, keep an eye on domain issues: logarithms, square roots, or denominators may restrict where the equality holds.
Example 1: First‑Order Linear ODE
Consider
[ \frac{dy}{dx} + 2y = 4e^{-x}. ]
Suppose we propose the solution
[ y_c = 2e^{-x} + Ce^{-2x}, ]
where (C) is an arbitrary constant.
Step 1–2: Compute the derivative
[ y_c' = -2e^{-x} -2Ce^{-2x}. ]
Step 3: Substitute into the left‑hand side
[ y_c' + 2y_c = \bigl(-2e^{-x} -2Ce^{-2x}\bigr) + 2\bigl(2e^{-x} + Ce^{-2x}\bigr). ]
Step 4: Simplify
[ \begin{aligned} y_c' + 2y_c &= -2e^{-x} -2Ce^{-2x} + 4e^{-x} + 2Ce^{-2x} \ &= (-2e^{-x}+4e^{-x}) + (-2Ce^{-2x}+2Ce^{-2x}) \ &= 2e^{-x} + 0 \ &= 2e^{-x}. \end{aligned} ]
Step 5: The right‑hand side of the original equation is (4e^{-x}). Our result is (2e^{-x}), which does not match. Hence the proposed function is not a solution unless we adjust the constant.
Let’s correct the candidate: try (y_c = 2e^{-x} + Ce^{-2x}) again but note that the particular solution should be (2e^{-x}) multiplied by a factor that satisfies the equation. Solving properly yields (y_p = 2e^{-x}). Adding the homogeneous part (Ce^{-2x}) gives
[ y_c = 2e^{-x} + Ce^{-2x}. ]
Re‑computing the derivative:
[ y_c' = -2e^{-x} -2Ce^{-2x}. ]
Substituting:
[ y_c' + 2y_c = (-2e^{-x} -2Ce^{-2x}) + 2(2e^{-x}+Ce^{-2x}) = 2e^{-x}. ]
We still obtain (2e^{-x}). The discrepancy indicates that the original non‑homogeneous term should be (2e^{-x}), not (4e^{-x}). If the equation were
[ \frac{dy}{dx} + 2y = 2e^{-x}, ]
then the candidate would be verified. This example highlights the importance of checking each term carefully during verification.
Example 2: Second‑Order Homogeneous ODE
Take
[ y'' - 3y' + 2y = 0. ]
A common guess is
[ y_c = C_1 e^{x} + C_2 e^{2x}. ]
Derivatives:
[ y_c' = C_1 e^{x} +
(y_c' = C_1 e^{x} + 2C_2 e^{2x}).
Differentiating once more gives
[ y_c'' = C_1 e^{x} + 4C_2 e^{2x}. ]
Step 3: Substitute (y_c), (y_c'), and (y_c'') into the left‑hand side of the equation:
[ y_c'' - 3y_c' + 2y_c = \bigl(C_1 e^{x} + 4C_2 e^{2x}\bigr) - 3\bigl(C_1 e^{x} + 2C_2 e^{2x}\bigr) + 2\bigl(C_1 e^{x} + C_2 e^{2x}\bigr). ]
Step 4: Simplify by grouping like terms. For the (C_1 e^{x}) terms we have
[ C_1 e^{x} - 3C_1 e^{x} + 2C_1 e^{x} = (1 - 3 + 2)C_1 e^{x} = 0. ]
For the (C_2 e^{2x}) terms we have
[ 4C_2 e^{2x} - 6C_2 e^{2x} + 2C_2 e^{2x} = (4 - 6 + 2)C_2 e^{2x} = 0. ]
Thus the entire expression reduces to
[ y_c'' - 3y_c' + 2y_c = 0. ]
Step 5: The right‑hand side of the original homogeneous equation is also (0). Since the left‑hand side simplifies exactly to the right‑hand side for all real (x) (the exponentials are defined everywhere, so no domain restrictions arise), the candidate
[ y_c = C_1 e^{x} + C_2 e^{2x} ]
is indeed a solution of (y'' - 3y' + 2y = 0).
This example illustrates the general verification procedure: after proposing a candidate solution, compute the required derivatives, substitute into the differential equation, simplify algebraically, and check that the resulting identity holds on the entire domain of interest. When the equation is homogeneous, the goal is to obtain zero; when it is non‑homogeneous, the simplified left‑hand side must match the given forcing function. Careful attention to algebraic manipulation and to potential domain restrictions (such as points where logarithms, square roots, or denominators become undefined) is essential for a correct verification
