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Solve
  • a)
    y = (c1 + c2log⁡x)x + logx + 2
  • b)
    y = (c1 + c2log⁡x)x + ex + 2
  • c)
    y = (c1 + c2log⁡x)x + elogx + 2
  • d)
    y = (c1 + c2⁡x)log x + log x + 2
Correct answer is option 'A'. Can you explain this answer?
Verified Answer
Solvea)y = (c1 + c2logx)x + logx + 2b)y = (c1 + c2logx)x + ex + 2c)y =...
Explanation:
Linear differential equations with variable coefficients which can be reduced to linear differential equations with constant coefficients by suitable substitution.
Euler Cauchy's homogeneous linear equation:

This is a Cauchy’s homogenous linear equation
Put x = et
⇒ t = log x

Now, the equation becomes,
(D(D − 1) − D + 1)y = t ⇒ (D − 1)2y = t
A.E. is (D – 1)2 = 0
⇒ D = 1, 1
C.F. = (c1 + ct) et

= t + 2
Complete solution is
y = (c1 + c2 t) et + t + 2
Putting t = log x
⇒ y = (c1 + c2 log x) x + log x + 2
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Most Upvoted Answer
Solvea)y = (c1 + c2logx)x + logx + 2b)y = (c1 + c2logx)x + ex + 2c)y =...
Explanation:
Linear differential equations with variable coefficients which can be reduced to linear differential equations with constant coefficients by suitable substitution.
Euler Cauchy's homogeneous linear equation:

This is a Cauchy’s homogenous linear equation
Put x = et
⇒ t = log x

Now, the equation becomes,
(D(D − 1) − D + 1)y = t ⇒ (D − 1)2y = t
A.E. is (D – 1)2 = 0
⇒ D = 1, 1
C.F. = (c1 + ct) et

= t + 2
Complete solution is
y = (c1 + c2 t) et + t + 2
Putting t = log x
⇒ y = (c1 + c2 log x) x + log x + 2
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Solvea)y = (c1 + c2logx)x + logx + 2b)y = (c1 + c2logx)x + ex + 2c)y = (c1 + c2logx)x + exlogx+ 2d)y = (c1 + c2x)log x + log x+ 2Correct answer is option 'A'. Can you explain this answer?
Question Description
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