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Solve the following differential equation y^{2}dx + (xy + x^{2}) dy = 0
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Solution
y^{2}dx + (xy + x^{2}) dy = 0
∴ `y^2 + (xy + x^2) ("d"y)/("d"x)` = 0
∴ `(xy + x^2) ("d"y)/("d"x)` = − y^{2}
∴ `("d"y)/("d"x) = (y^2)/(xy + x^2)` .....(i)
Put y = tx ......(ii)
Differentiating w.r.t. x, we get
`("d"y)/("d"x) = "t" + x "dt"/("d"x)` ......(iii)
Substituting (ii) and (iii) in (i), we get
`"t" + x "dt"/("d"x) = ("t"^2x^2)/(x("t"x) + x^2)`
∴ `"t" + x "dt"/("d"x) = ("t"^2x^2)/(x^2("t" + 1))`
∴ `"t" + x "dt"/("d"x) = ("t"^2)/(1 + "t")`
∴ `x "dt"/("d"x) = ("t"^2)/(1 + "t")  "t"`
= `("t"^2  "t"  "t"^2)/(1 + "t")`
= `(2"t"^2  "t")/(1 + "t")`
∴ `(1 + "t")/("t"^2 + "t") "dt" =  ("d"x)/x`
Integrating on both sides, we get
`int (1 + "t")/(2"t"^2 + "t") "dt" =  int ("d"x)/x`
∴ `int ((2"t" + 1)  "t")/("t"(21"t" + 1)) "dt" = int ("d"x)/x`
∴ `int (1/"t"  1/(2"t" + 1)) "dt" = int ("d"x)/x`
∴ `int 1/"t" "dt"  1/2 int 2/(2"t" + 1) "dt" = int ("d"x)/x`
∴ `log "t"  1/2 log2"t" + 1` = − logx + log c
∴ `log y/x  1/2 log2(y/x) + 1` = − logx + log c
∴ `log y  log x  1/2 log(2y + x)/x` = − logx + log c
∴ `1/2 logy^2  1/2 log(2y + x)/x` = log c
∴ `1/2 logy^2/((2y + x)/x)` = log c
∴ `1/2 log(xy^2)/(2y + x)` = log c
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