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Find the Particular Solution of the Differential Equation: Y ( 1 + Log X ) D X D Y − X Log X = 0 When Y = E2 and X = E - Mathematics and Statistics

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Question

Find the particular solution of the differential equation:

`y(1+logx) dx/dy - xlogx = 0`

when y = e2 and x = e

Solution

Given equation is

`y(1 + logx) dx/dy -xlogx = 0`

`:. y(1+logx) dx/dy = xlogx`

`:. y(1+logx)dx = xlogx dy`

Separating the variables

`1/ydy = (1+logx)/(xlogx) dx`

Integrating, we have

`int1/y dy = int (1+logx)/(xlogx) dx`

`:.log|y| = log|xlogx|+logc`

`:. log|y| = log|cxlogx|`

∴ y = cx log x is the general solution

Given x = e,    y = e2

∴ e2 = c.e.log e

∴ `e^2 = c.e`

∴ c = e

∴y = ex.logx

  Is there an error in this question or solution?

APPEARS IN

 2017-2018 (March) (with solutions)
Question 6.2.2 | 4.00 marks
Solution Find the Particular Solution of the Differential Equation: Y ( 1 + Log X ) D X D Y − X Log X = 0 When Y = E2 and X = E Concept: Methods of Solving First Order, First Degree Differential Equations - Differential Equations with Variables Separable.
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