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Exercise 6.3 · Q40

Q.Solve: y3−dydx=x2dydxy^3-\dfrac{dy}{dx}=x^2\dfrac{dy}{dx}

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y3−dydx=x2dydxy^3-\dfrac{dy}{dx}=x^2\dfrac{dy}{dx} gives y3=(x2+1)dydxy^3=(x^2+1)\dfrac{dy}{dx}, so dyy3=dxx2+1\dfrac{dy}{y^3}=\dfrac{dx}{x^2+1}. Integrating: $-\dfrac{1}{2y^ …

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