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Worked Examples · Example 2

Q.Verify that the function y=aebxy=ae^{bx} is a solution of the differential equation d2ydx2−b2y=0\frac{d^2y}{dx^2}-b^2y=0

Yanam CbseNCERTSubjective· 3mImportance★★★★★
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✓ Free question

Differentiating y=aebxy=ae^{bx} twice gives y′′=b2yy''=b^2y, which satisfies d2ydx2−b2y=0\dfrac{d^2y}{dx^2}-b^2y=0.

To verify a solution: substitute yy and its derivatives into the differential equation and check LHS == RHS. Here ddxebx=bebx\dfrac{d}{dx}e^{bx}=be^{bx}.

Given: y=aebxy=ae^{bx}; DE: d2ydx2−b2y=0\dfrac{d^2y}{dx^2}-b^2y=0.

  1. First derivative: dydx=a⋅bebx=abebx\dfrac{dy}{dx}=a\cdot be^{bx}=abe^{bx}.
  2. Second derivative: d2ydx2=ab⋅bebx=ab2ebx\dfrac{d^2y}{dx^2}=ab\cdot be^{bx}=ab^2e^{bx}.
  3. Note ab2ebx=b2 (aebx)=b2yab^2e^{bx}=b^2\,(ae^{bx})=b^2y.
  4. Substitute: d2ydx2−b2y=b2y−b2y=0\dfrac{d^2y}{dx^2}-b^2y=b^2y-b^2y=0.

LHS =0==0= RHS, so y=aebxy=ae^{bx} is a solution.

✓Final answer

d2ydx2−b2y=0\dfrac{d^2y}{dx^2}-b^2y=0 is satisfied — y=aebxy=ae^{bx} is verified as a solution.

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