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Question
Solve the following differential equation.
(x2 − y2 ) dx + 2xy dy = 0
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Solution
(x2 − y2 ) dx + 2xy dy = 0
∴ 2xy dy = (y2 - x2) dx
∴ `dy/dx = (y^2 - x^2)/(2xy) ......(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^2)/(2tx^2)`
∴ `x dt/dx = (t^2 - 1)/(2t )- t = (-(1+t^2))/(2t)`
∴ `2t/(1+t^2) dt = - dx/x`
Integrating on both sides, we get
`int 2t/(1+t^2) dt = - int dx/x`
∴ log |1 + t2| = -log |x| + log |c|
∴`log | 1+y^2/x^2| = log |c/x|`
∴ `(x^2 + y^2)/x^2 = c/x`
∴ x2 + y2 = cx
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