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प्रश्न
y2 dx + (xy + x2)dy = 0
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उत्तर
y2 dx + (xy + x2)dy = 0
∴ (xy + x2 ) dy = -y2 dx
∴ `dy/dx = -y^2/(xy + x^2)` ...(i)
Put y = tx ...(ii)
Differentiating w.r.t. x, we get
`dy/dx = t + x dt/dx` ...(iii)
Substituting (ii) and (iii) in (i), we get
∴ `t + x dt/dx = (-t^2 x^2)/(x.tx + x^2)`
∴ `t + x dt/dx = (-t^2 x^2)/(x^2(t+1)`
∴ `x dt/dx = (-t^2)/(t+1) -t`
∴ `x dt/dx = (-t^2 - t^2 - t)/(t+1)`
∴ `x dt/dx = (- (2t^2 + t))/(t+1)`
∴ `(t+1)/(2t^2 +t) dt = -1/x dx`
Integrating on both sides, we get
`int (t+1)/(2t^2 + t) dt = - int 1/x dx`
∴ `int (2t +1 - t)/(t(2t+1)) dt = - int 1/x dx`
∴ `int 1/t dt - int 1/(2t + 1) dt = -int 1/x dx`
∴ `log | t | -1/ 2 log |2t + 1| = -log |x| + log |c|`
∴ 2log| t | -log |2t + 1| = -2log |x| + 2 log |c|
∴ `2log |y/x| -log |(2y)/ x +1|=- 2log |x| + 2 log |c|`
∴ 2log |y| - 2log |x| - log |2y + x| + log |x| = - 2log |x| + 2log |c|
∴ log |y2| + log |x| = log |c2 |+ log |2y + x|
∴ log |y2x| = log |c2(x + 2y)|
∴ xy2 = c2 (x + 2y)
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