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प्रश्न
Solve the following differential equation:
(x2 – y2)dx + 2xy dy = 0
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उत्तर
Differential equation is
(x2 – y2) dx + 2xy dy = 0
`\implies dy/dx = (-(x^2 - y^2))/(2xy)` ...(1)
Which is a Homogeneous differential equation then put y = Vx in (1)
By (1) `d/dx (Vx) = ((Vx)^2 - x^2)/(2x(Vx))`
`\implies V + x (dV)/dx = (V^2 - 1)/(2V)`
`\implies x (dV)/dx = (V^2 - 1)/(2V) - V`
`\implies x (dV)/dx = (V^2 - 1 - 2V^2)/(2V)`
`\implies x (dV)/dx = (-V^2 - 1)/(2V)`
`\implies (2V)/(V^2 + 1) dV = - dx/x`
Now integrating both sides
`int (2V)/(V^2 + 1)dV = -int dx/x`
`\implies` log (V2 + 1) = – log x + log c
`\implies log(V^2 + 1) = log(C/x)`
`\implies` V2 + 1 = `C/x`
`\implies y^2/x^2 + 1 = C/x`
`\implies` y2 + x2 = Cx.
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