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Question
Find the differential equation whose general solution is
x3 + y3 = 35ax.
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Solution
x3 + y3 = 35ax ...(i)
Differentiating w.r.t. x, we get
`3x^3 + 3y^3 dy/dx = 35a` ...(ii)
Substituting (ii) in (i), we get
`x^3 + y^3 = (3x^2 + 3y^2 dy/dx)x`
∴ `x^3 + y^3 = 3x^3 + 3x*y^2 dy/dx`
∴ `2x^3 - y^3 +3xy^2dy/dx =0`, which is the required differential equation.
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