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प्रश्न
Solve the following differential equation y2dx + (xy + x2) dy = 0
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उत्तर
y2dx + (xy + x2) dy = 0
∴ `y^2 + (xy + x^2) ("d"y)/("d"x)` = 0
∴ `(xy + x^2) ("d"y)/("d"x)` = − y2
∴ `("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 log|2"t" + 1|` = − log|x| + log |c|
∴ `log |y/x| - 1/2 log|2(y/x) + 1|` = − log|x| + log |c|
∴ `log |y| - log |x| - 1/2 log|(2y + x)/x|` = − log|x| + log |c|
∴ `1/2 log|y^2| - 1/2 log|(2y + x)/x|` = log |c|
∴ `1/2 log|y^2/((2y + x)/x)|` = log |c|
∴ `1/2 log|(xy^2)/(2y + x)|` = log |c|
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