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प्रश्न
Solve x2ydx – (x3 + y3) dy = 0
उत्तर
x2ydx – (x3 + y3) dy = 0
x2ydx = (x3 + y3) dy
`("d"y)/("d"x) = (x^2y)/((x^3 + y^3))` ......(1)
This is a homogeneous differential equation, same degree in x and y
Let y = vx and `("d"y)/("d"x) = "v" + x "v"/("d"x)`
Equation (1)
⇒ `"v" + x "dv"/("d"x) = (x^2("v"x))/((x^3 + "v"^3x^3))`
⇒ `"v" + x "dv"/("d"x) = (x^3"v")/(x^3(1 + "v"^3))`
`"v" + x "dv"/("d"x) = "v"/((1+ "v"^3))`
`x "dv"/("d"x) = "v"/((1 + "v"^3)) - "v"`
`x "dv"/("d"x) = ("v" - "v"(1 + "v"^3))/((1 + "v"^3))`
`x "dv"/("d"x) = ("v" - "v" - "v"^4)/((1 + "v"^3)) = (-"v"^4)/((1 + "v"^3))`
`((1 +"v"^3))/"v"^4 "dv" = - 1/x "d"x`
Integrating on both sides
`int (1/"v"^4 + "v"^3/"v"^4) "dv" = - int 1/x "d"x`
`int "v"^-4 "dv" + int 1/"v" "dv" = - log x +"c"`
`[("v"^(-4 + 1))/(-4 + 1)] + log "v" + log x = "c"`
`["v"^-3/(-3)] + log "v"x + "c"`
`-1/(3"v"^3) + log (y/x) + "c"`
`x^3/(-3y^3) + log y` = c
⇒ logy = `x^3/(3y^3) + "c"`
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