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Exercise 10.7 · Q12

Q.dydx=sin⁡2x1+x3−3x21+x3y\dfrac{dy}{dx}=\dfrac{\sin^2x}{1+x^3}-\dfrac{3x^2}{1+x^3}y

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Rearrange to standard form, get a clean polynomial I.F., then integrate sin⁡2x\sin^2x using the half-angle identity.

Step 1. Rearrange. dydx+3x21+x3y=sin⁡2x1+x3\dfrac{dy}{dx}+\dfrac{3x^2}{1+x^3}y=\dfrac{\sin^2x}{1+x^3}. P=3x21+x3, Q=sin⁡2x1+x3P=\dfrac{3x^2}{1+x^3},\ Q=\dfrac{\sin^2x}{1+x^3}.

Step 2. Integrating factor. ∫P dx=ln⁡(1+x3)⇒I.F.=1+x3\int P\,dx=\ln\left(1+x^3\right)\Rightarrow I.F.=1+x^3. …

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