Question

Solve the following homogeneous differential equation:

An electric manufacturing company makes small household switches. The company estimates the marginal revenue function for these switches to be (x² + y²) dy = xy dx where x represents the number of units (in thousands). What is the total revenue function?

Answer 1

Marginal revenue for the switches

(x² + y²) dy = xy dx

`("d"y)/("d"x) = (xy)/((x^2 + y^2))`  .......(1)

It is a homogeneous differential equation same degree in x and y.

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

Equation (1) 

⇒ `"v" + x "dv"/("d"x) = (x("v"x))/(x^2 + "v"^2:n:^2) = (x^2("v"))/(x^2(1 + "v"^2))`

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

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

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

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

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

Integrating on both sides

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

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

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

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

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

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

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

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

`(-1)/(2 y^2/x^2) = log ("c"/((y/x)(x)))`

`(-x^2)/(2y^2) = log ("c"/y)`

⇒ `"c"/y = "e"^((-x^2)/(2y^2)`

`"ce"^(x^2/(2y^2))` = y

The total revenue function is 

⇒ y = `"ce"^(x^2/(2y^2))`