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Exercise 10.7 · Q8

Q.dydx+y(1−x)x=1−x\dfrac{dy}{dx}+\dfrac{y}{(1-x)\sqrt{x}}=1-\sqrt{x}

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The integrating factor here needs the substitution x=u2x=u^2 to evaluate; once found, the right side simplifies dramatically because Q⋅I.F.=1+xQ\cdot I.F.=1+\sqrt x.

Step 1. Identify P,QP,Q. P=1(1−x)x, Q=1−xP=\dfrac{1}{(1-x)\sqrt x},\ Q=1-\sqrt x.

Step 2. Integrate PP via x=u2, dx=2u du, x=ux=u^2,\ dx=2u\,du,\ \sqrt x=u. ∫2u du(1−u2)u=∫2 du1−u2=ln⁡∣1+u1−u∣=ln⁡∣1+x1−x∣\displaystyle\int\dfrac{2u\,du}{(1-u^2)u}=\int\dfrac{2\,du}{1-u^2}=\ln\left|\dfrac{1+u}{1-u}\right|=\ln\left|\dfrac{1+\sqrt x}{1-\sqrt x}\right|.

Step 3. Integrating factor. I.F.=1+x1−xI.F.=\dfrac{1+\sqrt x}{1-\sqrt x}. …

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