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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

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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`.

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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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