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Question
x2y dx – (x3 + y3) dy = 0
Solution
x2y dx – (x3 + y3) dy = 0
∴ x2y dx – (x3 + y3) = dy
∴ `dy/dx = (x^2y)/(x^3 + y^3)` …(i)
Put y = tx …(ii)
Differentiating w.r.t. x, we get
`dy/dx = t + x dt/dx` …(iii)
Substituting (ii) and (iii) in (i), we get
`t + x dt/dx = (x^2 . tx)/(x^3 + t^3 x^3)`
∴ `t + x dt/dx = (x^3.t)/(x^3(1+t^3))`
∴ `x dt/dx = t/(1+t^3) - t`
∴ `x dt/dx = (t-t-t^4)/(1+t^3)`
∴ `x dt/dx = (-t^4)/(1+t^3)`
∴ `(1+t^3)/t^4dt = - dx/x`
Integrating on both sides, we get
`int(1+t^3)/t^4 dt = - int 1/x dx `
∴ `int (1/t^4 + 1/t) dt = - int 1/x dx`
∴ `int t^-4 dt + int 1/t dt = - int1/x dx`
∴ `t^3/-3 + log |t| = - log |x| + c`
∴ `-1/(3t^3)+ log | t | = - log |x| + c`
∴ `- 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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