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Exercise 6.2 · Q12

Q.Obtain the differential equation by eliminating the arbitrary constant: x3+y3=4axx^3+y^3=4ax

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✓ Free question

Differentiate x3+y3=4axx^3+y^3=4ax w.r.t. xx: 3x2+3y2dydx=4a3x^2+3y^2\dfrac{dy}{dx}=4a. Substitute this value of 4a4a back into the original relation: x3+y3=x(3x2+3y2dydx)=3x3+3xy2dydxx^3+y^3=x\left(3x^2+3y^2\dfrac{dy}{dx}\right)=3x^3+3xy^2\dfrac{dy}{dx}. So y3−2x3=3xy2dydxy^3-2x^3=3xy^2\dfrac{dy}{dx}.

✓Final answer

3xy2dydx=y3−2x33xy^2\dfrac{dy}{dx}=y^3-2x^3

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