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Solve the following differential equation: (x2 + 3xy + y2)dx - x2 dy = 0

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प्रश्न

Solve the following differential equation:

(x2 + 3xy + y2)dx - x2 dy = 0

योग
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उत्तर

(x2 + 3xy + y2)dx - x2 dy = 0

∴ x2 dy = (x2 + 3xy + y2)dx

∴ `"dy"/"dx" = ("x"^2 + 3"xy" + "y"^2)/"x"^2`  ....(1)

Put y = vx

∴ `"dy"/"dx" = "v + x" "dv"/"dx"`

∴ (1) becomes, `"v + x" "dv"/"dx" = ("x"^2 + 3"x" * "vx" + "v"^2"x"^2)/"x"^2`

∴ `"v + x" "dv"/"dx" = 1 + 3"v" + "v"^2`

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

∴ `1/("v + 1")^2 "dv" = 1/"x" "dx"`

Integrating, we get

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

∴ `("v + 1")^-1/-1 = log |"x"| + "c"_1`

∴ `- 1/("v + 1") = log |"x"| + "c"_1`

∴ `- 1/("y"/"x" + 1) = log |"x"| + "c"_1`

∴ `- "x"/("y + x") = log |"x"| + "c"_1`

∴ `log |"x"| + "x"/("x + y") = - "c"_1`

∴ `log |"x"| + "x"/("x + y") = "c"`, where c = - c1

This is the general solution.

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अध्याय 6: Differential Equations - Exercise 6.4 [पृष्ठ २०३]

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बालभारती Mathematics and Statistics 2 (Arts and Science) [English] Standard 12 Maharashtra State Board
अध्याय 6 Differential Equations
Exercise 6.4 | Q 14 | पृष्ठ २०३

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