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Sum

**Solve the following differential equation:**

`("x" + 2"y"^3) "dy"/"dx" = "y"`

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#### Solution

`("x" + 2"y"^3) "dy"/"dx" = "y"`

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

∴ `"x"/"y" + 2"y"^2 = "dx"/"dy"`

∴ `"dy"/"dx" - 1/"y" * "x" = 2"y"^2` .....(1)

This is the linear differential equation of the form

`"dx"/"dy" + "P"*"x" = "Q"`, where P = `- 1/"y"` and Q = 2y^{2}

∴ I.F. = `"e"^(int "Pdy") = "e"^(int - 1/"y""dy")`

∴ = `"e"^(- log "y") = "e"^(log (1/"y")) = 1/"y"`

∴ the solution of (1) is given by

∴ `"x" * ("I.F.") = int "Q" ("I.F.") "dy" + "c"`

∴ `"x"(1/"y") = int 2"y"^2 xx 1/"y" "dy" + "c"`

∴ `"x"/"y" = 2 int "y" "dx" + "c"`

∴ `"x"/"y" = 2 * "y"^2/2 + "c"`

∴ x = y(c + y^{2})

This is the general solution.

Concept: Methods of Solving First Order, First Degree Differential Equations - Linear Differential Equations

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