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प्रश्न
Solve
`y log y dy/dx + x – log y = 0`
उत्तर
`y log y dy/dx + x – log y = 0`
∴ `dx/dy + 1/(y log y) x= 1/y`
The given equation is of the form `dx/dy + px = Q`
where, `P = 1 /(y logy )and Q = 1/y`
∴ `I.F. = e ^(int^pdy) = e^(int^(1/(ylogy)) dy) = e ^(log | log y|) = log y`
∴ Solution of the given equation is
`x(I.F.) =int Q (I.F.) dy + c_1`
∴ `x.log y = int 1/y log y dy + c_1`
In R. H. S., put log y = t
Differentiating w.r.t. x, we get
`1/y dy = dt`
∴ `x log y =int t dt + c_1 = t^2/2 + c_1`
∴`x log y =(logy)^2/2 + c_1`
∴ 2x log y = (log y)2 + c …[2c1 = c]
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