Q.Verify that the function y=aebx is a solution of the differential equation dx2d2y−b2y=0
Puducherry CbseNCERTSubjective· 3mImportance★★★★★
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Concept understanding — Solution Verification
Solution Verification — The Intuition
When you solve an equation, you perform operations on both sides to isolate the variable. But here's the catch: some operations are reversible (like adding 5 or multiplying by 3) and some are not (like squaring both sides or multiplying by an expression that could be zero). When you use a non-reversible step, you might introduce extra "solutions" that don't actually satisfy the original equation. These are called extraneous solutions.
Solution verification is the simple act of plugging your final answers back into the original equation to check which ones actually work. It's your safety net against these fake solutions.
Watch out
Never check against a simplified version of the equation — only the original one. A simplified version might have already lost information (like a denominator that was cancelled), so it could still accept an extraneous solution.
The Precise Statement
Definition:
Given an equation E(x)=0 and a set of candidate solutions S={x1,x2,…,xn} obtained by algebraic manipulation, solution verification is the process of substituting each xi into the original equation E(x)=0 and retaining only those xi for which the equality holds true.
Why it's necessary:
If at any step you performed an operation that is not bijective (one-to-one and onto) on the domain of the equation, the solution set of the transformed equation may be a superset of the solution set of the original equation. Common culprits:
Squaring both sides: x=2⟹x2=4, but x2=4 also gives x=−2, which is extraneous.
Multiplying by an expression that could be zero: x−1x=2⟹x=2(x−1) loses the restriction x=1.
Taking logarithms or exponentials without domain checks.
A Worked Example
Solve: x+6=x
Step 1 — Square both sides:
(x+6)2=x2⟹x+6=x2⟹x2−x−6=0⟹(x−3)(x+2)=0
So x=3 or x=−2.
Step 2 — Verify in the original equation:
For x=3: 3+6=9=3, and RHS is 3. Works.
For x=−2: −2+6=4=2, but RHS is −2. 2=−2. Extraneous.
Final answer:x=3 only.
Tip
Squaring is the most common source of extraneous roots. Whenever you square an equation, always verify. The same applies to raising both sides to any even power.
When You Can Skip Verification
You do not need to verify if every step you performed was a reversible transformation on the entire domain of the original equation. These include:
Adding or subtracting any expression (constant or variable)
Multiplying or dividing by a non-zero constant
Applying a strictly monotonic function (like ex, logx, or an odd power like x3)
But in practice, for Indian exams (JEE, board exams), always verify unless the problem explicitly says "without checking" or the solution is trivial. It costs 10 seconds and saves marks.
Important
Solution verification is not optional — it is part of the solution. In many exam marking schemes, presenting an extraneous root without discarding it loses marks. Always write: "On verification, x=__ satisfies the original equation, while x=__ does not."
Verifying this solution means differentiating y=aebx twice and checking that the resulting dx2d2y equals b2y.
✓Final answer
With y=aebx, y′′=b2aebx=b2y, so y′′−b2y=0. Verified.
Differentiating y=aebx twice gives y′′=b2y, which satisfies dx2d2y−b2y=0.
To verify a solution: substitute y and its derivatives into the differential equation and check LHS = RHS. Here dxdebx=bebx.
Given:y=aebx; DE: dx2d2y−b2y=0.
First derivative: dxdy=a⋅bebx=abebx.
Second derivative: dx2d2y=ab⋅bebx=ab2ebx.
Note ab2ebx=b2(aebx)=b2y.
Substitute: dx2d2y−b2y=b2y−b2y=0.
LHS =0= RHS, so y=aebx is a solution.
✓Final answer
dx2d2y−b2y=0 is satisfied — y=aebx is verified as a solution.