Question

Solve the following homogeneous differential equation:

The slope of the tangent to a curve at any point (x, y) on it is given by (y³ – 2yx²) dx + (2xy² – x³) dy = 0 and the curve passes through (1, 2). Find the equation of the curve

Answer 1

Given the slope is (y³ – 2yx²) dx + (2xy² – x³) dy = 0

(y³ – 2yx²) dx =  – (2xy² – x³) dy

`("d"y)/("d"x) = (y^3 - 2yx^2)/(x^3 - 2xy^2)`

This is a homogeneous differential equation

Put y = vx and `("d"y)/("d"x) = "v" + x "dv"/("d"x)`

`"v" + x "dv"/("d"x) = ("v"^3x^3 - 2"v"x^3)/(x^3 - 2x^3"v"^2) = ("v"^3 - 2"v")/(1 - 2"v"^2)`

`x "dv"/("d"x)  = ("v"^3 - 2"v")/(1 - 2"v"^2) - "v"`

= `("v"^3 - 2"v" - "v" + 2"v"^3)/(1 - 2"v"^2)`

= `(3"v"^3 - 3"v")/(1 - 2"v"^2)`

Separating the variables

`(1 - 2"v"^2)/(3("v"^3 - "v")) "dv" = ("d"x)/x`

`1/3 int (1 - 2"v"^2)/("v"("v"^2 - 1)) "dv" = int ("d"x)/x`

Using partial fraction method,

`(1 - 2"v"^2)/("v"("v"^2 - 1)) = (1 - 2"v"^2)/("v"("v" + 1)("v" - 1))`

= `"A"/"v" + "B"/("v" + 1) + "C"/("v" - 1)`

`1 - 2"v"^2 = "A"("v" + 1)("v" - 1) + "Bv"("v" - 1) + "Cv"("v" + 1)`

Put, v = 0 ⇒ A = – 1

v = 1 ⇒ C = `-1/2`

v =  – 1 ⇒ B = `- 1/2`

`1/3 [int (-1)/"v" "dv" - int 1/(2("v" + 1)) "dv" - 1/2 int 1/("v" - 1) "dv"] = int ("d"x)/x`

`1/3 [- log"v" - 1/2 log ("v" + 1) - 1/2 log("v" - 1)] = log x + log "c"`

`- 1/3 [log "v" + logsqrt("v" + 1) + logsqrt("v" - 1)] = log "c"x`

`- 1/3 log "v"sqrt("v"^2 - 1) = log "c"x`

⇒ `1/(("v"sqrt("v"^2 - 1))^(1/3)` = cx

`1/(("v"sqrt("v"^2 - 1))` = c³x³

Replace, v = `y/x`

`1/(y/xsqrt(y^2/x^2) - 1)` = c³x³

`1/(y/x sqrt((y^2 - x^2)/x^2)` = c³x³ 

⇒ `x^2/(ysqrt(y^2 - x^2)` = c³x³  

`ysqrt(y^2 - x^2) = 1/("c"^3x) = 1/("k"x)`

k = c³

The curve passes through (1, 2) (i.e) x = 1, y = 2

`2sqrt(3) = 1/"k"`

Thus, `ysqrt(y^2 - x^2) = (2sqrt(3))/x`

or

`xysqrt(y^2 - x^2) = 2sqrt(3)` is the required solution