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प्रश्न
Solve the following differential equation:
x2 dy + (xy + y2) dx = 0, when x = 1 and y = 1
उत्तर
x2 dy + (xy + y2) dx = 0
⇒ x2 dy = -(xy + y2 ) dx
⇒ `(dy)/(dx) = - ((xy + y^2))/x^2`
Let y = vx
`(dy)/(dx) = v + x (dv)/(dx)`
`v + x (dv)/(dx) = - ((vx^2 + v^2 x^2))/x^2`
`v + x (dv)/(dx) = - (v + v^2)`
` x (dv)/(dx) = - 2v - v^2`
⇒ `(dv)/(v^2 + 2v) = - (dx)/x`
⇒ `(dv)/(v^2 + 2v) + (dx)/x = 0`
⇒ ` int_ (dv)/((v + 1)^2 - (1)^2) + int_ (dx)/x = 0`
⇒ `(1)/(2) log |[ v+ 1 -1]/[v + 1 + 1]| + log x = c`
⇒ ` log |[ v]/[ v + 2]| + 2 log x = 2c`
⇒ ` log |[ vx^2]/[ v + 2]| = 2c`
⇒ ` |[ vx^2]/[ v + 2]| = e^2c`
⇒ ` ( vx^2)/( v + 2) = ± e^2""^c = "A" ("say")`
⇒ `(yx)/(y/2 + 2) = "A"`
⇒ `x^2y = "A" (y + 2x)`
Now, when y = 1, x = 1
⇒ 1 = A (3)
⇒ A = `(1)/(3)`
∴ `y + 2x = 3x^2y`
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