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प्रश्न
For the differential equation, find the general solution:
`x log x dy/dx + y= 2/x log x`
उत्तर
The given equation
`x log x dy/dx + y = 2/x log x`
or `dy/dx + y/(x log x) = 1/x^2` ....(i)
Comparing with `dy/dx + Py = Q`,
`P = 1/(x log x)` and `Q = 2/x^2`
∴ `I.F. = e^(intP dx) = e^(int 1/(x log x)dx)`
`= e^(log(log x)) = log x`
Hence the required solution
∴ `y × I.F. = int I.F. xx Q dx + C`
`=> y log x = 2 int 1/x^2 (log x) dx + C`
`=> y log x = 2 [log x (- 1/x) - int 1/x ((- 1)/x) dx] + C`
`=> y log x = (- 2)/x log x + 2 int 1/x^2 dx + C`
`=> y log x = (- 2)/x log x - 2/x + C`
⇒ `y = log x = (- 2)/x (1 + log |x|) + C`
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