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Exercise 6.5 · Q78

Q.Solve: (1−x2)dydx+2xy=x(1−x2)1/2(1-x^2)\dfrac{dy}{dx}+2xy=x(1-x^2)^{1/2}

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(1−x2)dydx+2xy=x(1−x2)1/2(1-x^2)\dfrac{dy}{dx}+2xy=x(1-x^2)^{1/2} rewrites as dydx+2x1−x2y=x1−x2\dfrac{dy}{dx}+\dfrac{2x}{1-x^2}y=\dfrac{x}{\sqrt{1-x^2}}, linear with P=2x1−x2, Q=x1−x2P=\dfrac{2x}{1-x^2},\ Q=\dfrac{x}{\sqrt{1-x^2}}. I.F. =e∫2x dx/(1−x2)=e−log⁡(1−x2)=11−x2=e^{\int 2x\,dx/(1-x^2)}=e^{-\log(1-x^2)}=\dfrac{1}{1-x^2}. So y1−x2=∫x1−x2⋅11−x2dx+c=∫x(1−x2)3/2dx+c\dfrac{y}{1-x^2}=\int\dfrac{x}{\sqrt{1-x^2}}\cdot\dfrac{1}{1-x^2}dx+c=\int\dfrac{x}{(1-x^2)^{3/2}}dx+c. With t=1−x2, dt=−2x dxt=1-x^2,\ dt=-2x\,dx: …

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