# Solve the Differential Equation : ( X 2 + 3 Xy + Y 2 ) D X − X 2 D Y = 0 Given that Y = 0 When X = 1 . - Mathematics

Sum

Solve the differential equation : ("x"^2 + 3"xy" + "y"^2)d"x" - "x"^2 d"y" = 0  "given that"  "y" = 0  "when"  "x" = 1.

#### Solution

Here ("x"^2 + 3"xy" + "y"^2)d"x" - "x"^2d"y" = 0

⇒ (d"y")/(d"x") = ("x"^2 + 3"xy"+"y"^2)/("x"^2)

⇒ (d"y")/(d"x") = 1 + 3("y")/("x") + ("y"^2)/("x"^2)

Put y = vx

⇒ (d"y")/(d"x") = "v" + "x"(d"v")/(d"x")

∴ "v" + "x"(d"v")/(d"x") = 1 + 3"v"+"v"^2

⇒ "x"(d"v")/(d"x") = 1 + 2"v"+"v"^2

⇒ int_  (d"v")/(("v"+1)^2) = int_  (d"x")/("x")

⇒ -(1)/(("v"+1)) = log|"x"| + "C"

⇒ -("x")/("y"+"x") = log|"x"|+"C"

As 1 y = 0 when x = 1 so, -(1)/(0+1) = log|1|+"C"

⇒ "C" = -1.

Hence the required solution is, -("x")/("y"+"x")   = log|"x"|-1

⇒ -"x" = "y" log|"x"| +"x" log |"x"| -"y" -"x"

∴ "y" = log|"x"| ("x" + "y")

or,

"y" = ("x" log|"x"|)/(1-log|"x"|).

Concept: General and Particular Solutions of a Differential Equation
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