Question

Solve the following differential equation:

`2xy"d"x + (x^2 + 2y^2)"d"y` = 0

Answer 1

The given differential equation can be written as

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

This is a homogenous differential equation

Putting y = vx

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

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

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

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

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

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

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

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

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

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

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

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

Multiply and  Divide by 3, we get

`1/3 int(3 + 6"v"^2)/(3"v" + "v"^3) "dv" = -int ("d"x)/x`

`1/3 log(3"v" + "v"^3) = - logx + log |"C"_1|`

`1/3 log(3"v" + 2"v"^3) + logx = log |"C"_1|`

log (3v + 2v³) + 3log (x) = 3 log (C₁)

log (3v + 2v³) + log (x)³ = log (C₁)³

log (3v + 2v³)x³ = log C₁³

(3v + 2v³)x³ = C₁³ 

`(3(y/x) + 2(y/x)^3)x^3` =  C₁³ 

`((3y)/x + (2y^3)/x^3)x^3` = C₁³ 

`((3x^2y + 2y^3)x^3)/x^3` = C₁³ 

3x²y + 2y³ = C₁³

3x²y + 2y³ = C is a required solution.