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प्रश्न
Find the particular solution of the differential equation:
`y(1+logx) dx/dy - xlogx = 0`
when y = e2 and x = e
उत्तर
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
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