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Question
Solve the following differential equation.
x2y dx − (x3 + y3) dy = 0
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Solution
Given: x2y dx − (x3 + y3) dy = 0
∴ x2y dx = (x3 + y3) dy
∴ `dy/dx = (x^2y)/(x^3+y^3)` ...(i)
Which is a homogeneous differential equation.
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^4 dt = -dx/x`
Which is in variable separable form.
Integrating on both sides, we get
∴ `int(1 + t^3)/t^4dt = - int 1/x dx`
∴ `int(1/t^4 + 1/t)dt = - int1/x dx`
∴ `int t^-4 dt + int 1/t dt = - int 1/x dx`
∴ ` t^-3/-3 + log |t| = - log |x| + c`
∴ `- 1/(3t^3) + log t = - log x + c`
Resubstituting the value of `t = y/x`, we get
∴ `-1/3 * 1/(y/x)^3 + log (y/x) = - log x + c`
∴ `-x^3/(3y^3) + log y - log x = c - log x`
∴ `log y - x^3/(3y^3) = c`
∴ `log y - x^3/(3y^3) = c` is the required general solution.
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