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प्रश्न
Solve (x2 + y2) dx + 2xy dy = 0
उत्तर
(x2 + y2) dx + 2xy dy = 0
2xy dy = – (x2 + y2) dx
`("d"y)/("d"x) = (-(x^2 + y^2))/(2xy)` ........(1)
This is a homogeneous differential equation
Put y = vx and `("d"y)/("d"x) = "v" + x "dv"/("d"x)`
∴ (1) ⇒ `"v" + x "dv"/("d"x) = (-(x^2 + "v"^2x^2))/(2x("v"x))`
`"v" + x "dv"/("d"x) = (-x^2(1 + "v"^2))/(2"v"x^2)`
`"v" + x "dv"/("d"x) = (-(1 + "v"^2))/(2"v")`
`x "dv"/("d"x) = (-(1 + "v"^2)/(2"v") - "v"`
`x "dv"/("d"x) = (-1 - "v"^2 - 2"v"^2)/(2"v")`
= `(-3"v"^2 - 1)/(2"v")`
= `(-(3"v"^2+ 1))/(2"v")`
`((2"v"))/((3"v"^2 + 1)) "dv" = - 1/x "d"x`
Integrating on both sides
`int ((2"v"))/((3"v"^2 + 1)) "dv" = - int 1/x "d"x`
`1/3 int (6"v")/((3"v"^2 + 1)) "dv" = - int 1/x "d"x`
`1/3 log (3"v"^2 + 1) = - log x + log "c"`
`log (3"v"^2 + 1)^(1/3) + log x = log "c"`
`log x (3"v"^2 + 1)^(1/3) = log "c"`
⇒ `x(3"v"^2 + 1)^(1/3)` = c
⇒ `x[(3y^2)/x^3 + 1]^(1/3)` = c
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