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Question
Solve the following differential equation
x2y dx – (x3 + y3)dy = 0
Solution
x2y dx – (x3 + y3)dy = 0
⇒ (x3 + y3).dy = x2y.dx
⇒ `(dy)/(dx) = (x^2y)/(x^3 + y^3)` ......(1)
Let y = vx ......(2)
∴ `(dy)/(dx) = "v" + x (d"v")/(dx)`
∴ Equation (1) can be written as
`"v" + x. (d"v")/(dx) = (x^2."v"x)/(x^3 + "v"^3x^3)`
= `(x^3."v")/(x^3(1 + "v"^3))`
⇒ `"v" + x. (d"v")/(dx) = "v"/(1 + "v"^3)`
⇒ `x. (d"v")/(dx) = "v"/(1 + "v"^3) - "v"`
= `("v" - "v" - "v"^4)/(1 + "v"^3)`
⇒ `x. (d"v")/(dx) = - "v"^4/(1 + "v"^3)`
⇒ `(1 + "v"^3)/"v"^4 . d"v" = - 1/x. dx`
Integrating both sides. we get
`int (1 + "v"^3)/"v"^4. d"v" = - int 1/x. dx`
⇒ `int (1/"v"^4 + 1/"v"). d"v" = - int 1/x. dx`
⇒ `int "v"^-4. d"v" + int 1/"v". d"v" = - int 1/x. dx`
⇒ `"v"^(-3)/(-3) + log |"v"| = - log |x| + C`
⇒ `- 1/3 . 1/"v"^3 + log |"v"| = - log |x| + C`
Resubstituting the value of v from (2),
⇒ `- 1/3 1/(y/x)^3 + log |y/x| = - log |x| + C`
⇒ `(-x^3)/(3y^3) + log |y| - log |x| = - log |x| + C`
∴ `log y = x^3/(3y^3) + C`
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