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प्रश्न
Solve the following differential equation:
`(x^3 + y^3)"d"y - x^2 y"d"x` = 0
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उत्तर
The given equation can be written as
`(x^3 + y^3)"d"y = x^2 y"d"x`
`("d"y)/("d"x) = (x^2y)/(x^3 + y^2)`
This is a homogeneous equation
y = vx
`("d"y)/("d"x) = "v" + x "dv"/("d"x)`
`"v" + x "dv"/("d"x) =(x^2 xx "v"x)/(x^3 + "v"^3x^3)`
= `(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"`
= `("v" - "v"(1 + "v"^3))/(1 + "v"^3)`
`x "dv"/("d"x) = ("v" - "v" - "v"^4)/(1 + "v"^3)`
`x "dv"/("d"x) = (-"v"^4)/(1 + "v"^3)`
`((1 + "v"^3))/"v"^4 "dv" = (-"dv")/x`
On integrating,
`int (1 + "v"^3)/"v"^4 "dv" = - int ("d"x)/x`
`int (1/"v"^4 + "v"^3/"v"^4) "dv" = - log x + log "c"`
`int 1/"v"^4 "dv" + int 1/"v" "dv" = - logx + log "c"`
`int "v"^-4 "dv" + log "v" = - logx + log "c"`
`("v"^(-4 + 1))/(4 + 1) + log "v" = - log x + log "c"`
`"v"^(-3)/(-3) + log "v" + log x = log "c"`
`v^(-3)/(-3) + log("v"x) = log "c"`
`(-1)/(3"v"^3) + log("v"x) = log "c"`
`log "c" + 1/(3(y/x)^3) = log[y/x x]`
∵ y = vx
v = `y/x`
`log y = log "c" + 1/((3y^3)/x^3`
`log y = log "c" + x^3/(3y^3)`
`log y - log "c" = x^3/(3y^3)`
`log(y/"c") = x^3/(3y^3)`
`y/"c" = "e"^(x^3/(3y^3))`
y = `"ce"^(x^3/(3y^3))`
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