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A differential equation is given as:
The solution of the differential equation in terms of arbitrary constants C1 and C2 is
The solution of the differential equation in terms of arbitrary constants C1 and C2 is
- a)y = C2x2 + C1x + 2
- b)
- c)y = C1x2 + C2x + 4
- d)
Correct answer is option 'A'. Can you explain this answer?
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Most Upvoted Answer
A differential equation is given as:The solution of the differential e...
Given differential equation:
is the standard form of Euler-Cauchy DE.
So, let x = ez
⇒ dx = ezdz = x.dz
Similarly, we can obtain
⇒ xD = θ
⇒ x2D2 = θ (θ - 1)
Making these substitutions in given DE, we get
(θ (θ - 1) – 2θ + 2) y = 4
θ (θ - 1) – 2(θ – 1) = 0
⇒ Solution for this will constitute of CF & PI.
⇒ CF = (θ – 1) (θ – 2) = 0 ⇒ θ = 1 & θ = 2
⇒ y = c1ez + c2e2z = c1x + c2x2
⇒ y = CF + PI = c2x2 + c1x + 2
{We can also solve this problem by differentiating option also, if we don’t remember the process.}
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A differential equation is given as:The solution of the differential equation in terms of arbitrary constants C1and C2isa)y = C2x2+ C1x + 2b)c)y = C1x2+ C2x + 4d)Correct answer is option 'A'. Can you explain this answer?
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