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प्रश्न
y dx – x dy + log x dx = 0
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उत्तर
y dx – x dy + log x dx = 0
y dx – x dy = - log x dx
Dividing throughout by dx, we get
`y-x dy/dx = – log x `
∴ `-xdy/dx + y = - log x`
∴ `dy/dx - 1/(x y) = logx/x`
The given equation is of the form
`dy/dx + py = Q`
where, `P = -1/x and Q = logx/x`
∴ I.F. = `e ^(int^(pdx) = e^(int^(-1/xdx) e ^-logx`
= `e^(logx ^-1) = x ^-1 = 1/x`
∴ Solution of the given equation is
`y(I.F.) =int Q (I.F.) dx + c`
∴ `y/x = int logx/x xx1/xdx+c`
In R. H. S., put log x = t …(i)
∴ x = et
Differentiating (i) w.r.t. x, we get
`1/xdx = dt`
∴ `y/x = int t/e^t dt +c`
∴ `y/x = int te^t dt +c`
= `t int e^-t dt - int (d/dt(t)xxint e^-t dt) dt +c `
= `-te^-t - int (-e^-t) dt +c`
= `-te^-t + int e^-t dt +c`
= – te–t – e –t + c
= `(-t-t)/e^t + c`
= `(- logx -1)/x +c`
∴ y = cx – (1 + log x)
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