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Question
Solve the following differential equation:
(9x + 5y) dy + (15x + 11y)dx = 0
Solution
(9x + 5y) dy + (15x + 11y)dx = 0
∴ (9x + 5y) dy = - (15x + 11y) dx
∴ `"dy"/"dx" = (- (15"x" + 11"y"))/(9"x" + 5"y")` ...(1)
Put y = vx
∴ `"dy"/"dx" = "v + x" "dv"/"dx"`
∴ (1) becomes, `"v + x" "dv"/"dx" = (- (15"x" + 11"y"))/(9"x" + 5"y")`
∴ `"v + x" "dv"/"dx" = (- (15 + 11"v"))/(9 + 5"v")`
∴ `"v + x" "dv"/"dx" = (- (15 + 11"v"))/(9 + 5"v") - "v" = (- 15 - 11"v" - 9"v" - 5"v"^2)/(9 + "5v")`
∴ `"x" "dv"/"dx" = (- 5"v"^2 - 20"v" - 15)/(9 + "5v") = - ((5"v"^2 + 20"v" + 15)/(5"v" + 9))`
∴ `("5v + 9")/("5v"^2 + 20"v" + 15) "dv" = - 1/"x" "dx"` ....(2)
Integrating, we get
`1/5 int ("5v" + 9)/("v"^2 + 4"v" + 3) "dv" = - int 1/"x" "dx"`
Let `("5v" + 9)/("v"^2 + 4"v" + 3) = ("5v" + 9)/(("v" + 3)("v" + 1)) = "A"/("v + 3") + "B"/("v + 1")`
∴ 5v + 9 = A(v + 1) + B(v + 3)
Put v + 3 = 0, i.e. v = - 3, we gwr
- 15 + 9 = A(- 2) + B(0)
∴ - 6 = - 2A
∴ A = 3
Put v + 1 = 0, i.e. v = - 1, we get
- 5 + 9 = A(0) + B(2)
∴ 4 = 2B
∴ B = 2
∴ `("5v" + 9)/("v"^2 + 4"v" + 3) = 3/("v + 3") + 2/("v + 1")`
∴ (2) becomes,
`1/5 int (3/("v + 3") + 2/("v + 1"))"dv" = - int 1/"x" "dx"`
∴ `3/5 int 1/("v + 3")"dv" + 2/5 int1/("v + 1") "dv" = - int 1/"x" "dx"`
∴ `3/5 log |"v + 3"| + 2/5 log |"v + 1"| = - log |"x"| + "c"_`
∴ 3 log |v + 3| + 2 log |v + 1| = - 5 log x + 5c1
∴ `log |("v + 3")^3| + log |("v" + 1)^2| = - log |"x"^5| + log "c",` where 5c1 = log c
∴ `log |("v + 3")^3 ("v + 1")^2| = log |"c"/"x"^5|`
∴ `("v + 3")^3 ("v + 1")^2 = "c"/"x"^5`
∴ `("y"/"x" + 3)^3 ("y"/"x" + 1)^2 = "c"/"x"^5`
∴ `("x + y")^2 (3"x" + "y")^3 = "c"`
This is the general solution.
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