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Question
Solve the following differential equation:
`2xy"d"x + (x^2 + 2y^2)"d"y` = 0
Solution
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.
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