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प्रश्न
Find the particular solution of the following differential equation:
(x + y)dy + (x - y)dx = 0; when x = 1 = y
उत्तर
(x + y)dy + (x - y)dx = 0
∴ (x + y)dy = - (x - y)dx
∴ `"dy"/"dx" = ("y - x")/("y + x")` ...(1)
Put y = vx ∴ `"dy"/"dx" = "v" + "x" "dv"/"dx"`
∴ (1) becomes, `"v" + "x" "dv"/"dx" = ("vx" - "x")/("vx" + "x") = ("v - 1")/("v + 1")`
∴ `"x" "dv"/"dx" = ("v - 1")/("v + 1") - "v" = ("v" - 1 - "v"^2 - "v")/("v + 1")`
∴ `"x" "dv"/"dx" = - ((1 + "v"^2)/(1 + "v"))`
∴ `(1 + "v")/(1 + "v"^2) "dv" = - 1/"x" "dx"`
Integrating both sides, we get
`int (1 + "v")/(1 + "v"^2) "dv" = - int 1/"x" "dx"`
∴ `int (1/(1 + "v"^2) + "v"/(1 + "v"^2))"dv" = - int 1/"x" "dx"`
∴ `int 1/(1 + "v"^2) "dv" + 1/2 int "2v"/(1 + "v"^2)"dv" = - int 1/"x" "dx"`
∴ `tan^-1 "v" + 1/2 log |1 + "v"^2| = - log "x" + "c"` .....`[because "d"/"dv" (1 + "v"^2) = 2"v" and int ("f"'(x))/("f"("x")) "dv" = log |"f"("v")| + "c"]`
∴ `tan^-1 ("y"/"x") + 1/2 log |1 + "y"^2/"x"^2| = log |"x"| + "c"`
∴ `tan^-1 ("y"/"x") + 1/2 log |("x"^2 + "y"^2)/"x"^2| = - log |"x"| + "c"`
∴ `tan^-1 ("y"/"x") + 1/2 log "x"^2 + "y"^2 - 1/2 log |"x"^2| = - log |"x"| + "c"`
∴ `tan^-1 ("y"/"x") + log sqrt("x"^2 + "y"^2) - log |"x"| = - log |"x"| + "c"`
∴ `tan^-1 ("y"/"x") + log sqrt("x"^2 + "y"^2) = "c"`
This is the general solution.
When x = 1 = y, we have
`tan^-1 (1) + log sqrt(1^2 + 1^2) = "c"`
∴ `tan^-1 (tan pi/4) + log sqrt 2 = "c"`
∴ c = `pi/4 + log sqrt2`
∴ the particular solution is
∴ `tan^-1 ("y"/"x") + log sqrt("x"^2 + "y"^2) = pi/4 + log sqrt2`
Notes
The answer in the textbook is incorrect.
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