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

Sum

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

Concept: Methods of Solving First Order, First Degree Differential Equations - Differential Equations with Variables Separable Method
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