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प्रश्न
Solve the following differential equation.
`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/xdx`
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_1|`
∴ log |1 + t2 | = 2 log |x| + 2log |c1|
= log |x2| + log |c| …[logc12 = log c]
∴ log |1 + t2| = log |cx 2|
∴ 1 + t2 = cx2
∴ `1+ y^2/x^2 = cx^2`
∴ x2 + y2 = cx4
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