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प्रश्न
`xy dy/dx = x^2 + 2y^2`
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उत्तर
`xy dy/dx = x^2 + 2y^2`
∴ `dy/dx = x^2 + (2y^2)/ (xy)` …(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 = (x^2 + 2t^2 x^2)/ (x(tx))`
∴ `t + x dt/dx = (x^2 (1+2t^2))/(x^2t)`
∴ `x dt/dx (1+2t^2)/t - t = (1+ t^2)/t`
∴ `t/(1+t^2) dt = 1/x dx`
Integrating on both sides, we get
`1/2 int (2t)/(1+t^2) dt = int dx/x`
∴ `1/2 log|1 + t^2| = log |x| + log |c|`
∴ log |1 + t2 | = 2 log |x| + 2log |c|
= log |x2 | + log |c2|
∴ log |1 + t2 | = log |c2 x2|
∴ 1 + t2 = c2x2
∴ `1 + y^2/x^2 = c^2x^2`
∴ x2 + y2 = c2 x4
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