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NCERT Exemplar · Q92

Q.Which of the following is the general solution of d2ydx2−2dydx+y=0\frac{d^2y}{dx^2}-2\frac{dy}{dx}+y=0?
(A) y=(Ax+B)exy=(Ax+B)e^x
(B) y=(Ax+B)e−xy=(Ax+B)e^{-x}
(C) y=Aex+Be−xy=Ae^x+Be^{-x}
(D) y=Acos⁡x+Bsin⁡xy=A\cos x+B\sin x

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Substituting the candidate y=(Ax+B)exy=(Ax+B)e^x into y′′−2y′+yy''-2y'+y makes the left side collapse to 00 for every xx; since it also carries the two arbitrary constants A,BA,B needed for a second-order equation, option (A) is the general solution.

This is a multiple-choice question, so the fastest correct route is verification: we don't have to derive the answer from scratch — we just substitute the given candidate function into the differential equation and check that it makes both sides equal (an identity in xx). A general solution of a second-order equation must satisfy the equation and contain two independent arbitrary constants; option (A), y=(Ax+B)exy=(Ax+B)e^x, has the constants AA and BB, so it's the natural one to test.

Step 1 — Compute the derivatives that appear in the equation.

Start from

y=(Ax+B)ex.y=(Ax+B)e^x.

Using the product rule, ddx[(Ax+B)ex]=A ex+(Ax+B)ex\dfrac{d}{dx}\big[(Ax+B)e^x\big]=A\,e^x+(Ax+B)e^x, so

y′=(Ax+A+B)ex.y'=(Ax+A+B)e^x.

Differentiate once more, again by the product rule:

y′′=A ex+(Ax+A+B)ex=(Ax+2A+B)ex.y''=A\,e^x+(Ax+A+B)e^x=(Ax+2A+B)e^x.

Step 2 — Substitute yy, y′y', y′′y'' into the left-hand side.

y′′−2y′+y=(Ax+2A+B)ex−2(Ax+A+B)ex+(Ax+B)ex.y''-2y'+y=(Ax+2A+B)e^x-2(Ax+A+B)e^x+(Ax+B)e^x.

Factor out the common exe^x:

=[(Ax+2A+B)−2(Ax+A+B)+(Ax+B)]ex.=\big[(Ax+2A+B)-2(Ax+A+B)+(Ax+B)\big]e^x.

Step 3 — Simplify the bracket.

Collect like terms:

Ax−2Ax+Ax=0,2A−2A=0,B−2B+B=0.Ax-2Ax+Ax=0,\qquad 2A-2A=0,\qquad B-2B+B=0.

So the bracket is 00, and

y′′−2y′+y=0⋅ex=0.y''-2y'+y=0\cdot e^x=0. …

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