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Exercise 6.3 · Q50

Q.Reduce to variable separable form and solve: (x−y)2dydx=a2(x-y)^2\dfrac{dy}{dx}=a^2

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Put u=x−yu=x-y, so dudx=1−dydx\dfrac{du}{dx}=1-\dfrac{dy}{dx}, i.e. dydx=1−dudx\dfrac{dy}{dx}=1-\dfrac{du}{dx}. The equation (x−y)2dydx=a2(x-y)^2\dfrac{dy}{dx}=a^2 becomes u2(1−dudx)=a2u^2\left(1-\dfrac{du}{dx}\right)=a^2, i.e. u2dudx=u2−a2u^2\dfrac{du}{dx}=u^2-a^2, so u2u2−a2du=dx\dfrac{u^2}{u^2-a^2}du=dx. Since u2u2−a2=1+a2u2−a2\dfrac{u^2}{u^2-a^2}=1+\dfrac{a^2}{u^2-a^2}, integrating gives u+a2log⁡∣u−au+a∣=x+cu+\dfrac{a}{2}\log\left|\dfrac{u-a}{u+a}\right|=x+c. Substitu …

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