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प्रश्न
For the following differential equation find the particular solution satisfying the given condition:
`y(1 + log x) dx/dy - x log x = 0, y = e^2,` when x = e
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उत्तर
`y(1 + log x) dx/dy - x log x` = 0
∴ `(1 + log x)/(x log x)dx - dy/y` = 0
Integrating both sides, we get
∴ `int (1 + log x)/(x log x)dx - dy/y` = c1 .....(1)
Put x log x = t
Then `[x * d/dx (log x) + (log x) * d/dx (x)]dx` = dt
∴ `[x/x + (log x)(1)]dx` = dt
∴ `(1 + log x)dx` = dt
∴ `int (1 + log x)/(x log x)dx = intdt/t = log |t| = log |x log x|`
∴ From (1), the general solution is
log |x log x| – log |y| = log c, where c1 = log c
∴ log `|(x log x)/y|` = log c
∴ `(x log x)/y` = c
∴ x log x = cy
This is the general solution.
Now, y = `"e"^2`, when x = e
∴ e log e = c.e2
∴ 1 = c.e ...[∵ log e = 1]
∴ c = `1/e`
∴ The particular solution is x log x = `(1/e)y`
∴ y = ex log x.
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