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Exercise 10.6 · Q7

Q.(1+3ey/x)dy+3ey/x(1−yx)dx=0\left(1+3e^{y/x}\right)dy+3e^{y/x}\left(1-\dfrac{y}{x}\right)dx=0, given that y=0y=0 when x=1x=1

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After the y=vxy=vx substitution, notice the numerator 1+3ev1+3e^v is exactly ddv(3ev+v)\frac{d}{dv}(3e^v+v) — the integral is then immediate, without partial fractions.

Step 1. Rewrite as dydx\dfrac{dy}{dx}. dydx=−3ey/x(1−y/x)1+3ey/x\dfrac{dy}{dx}=\dfrac{-3e^{y/x}\left(1-y/x\right)}{1+3e^{y/x}} — homogeneous of degree 00.

Step 2. Substitute y=vxy=vx. v+xdvdx=−3ev(1−v)1+3evv+x\dfrac{dv}{dx}=\dfrac{-3e^v(1-v)}{1+3e^v}.

Step 3. Isolate xdvdxx\dfrac{dv}{dx} and simplify the numerator. xdvdx=−3ev(1−v)−v(1+3ev)1+3ev=−3ev+3vev−v−3vev1+3ev=−(3ev+v)1+3evx\dfrac{dv}{dx}=\dfrac{-3e^v(1-v)-v(1+3e^v)}{1+3e^v}=\dfrac{-3e^v+3ve^v-v-3ve^v}{1+3e^v}=\dfrac{-(3e^v+v)}{1+3e^v}.

Step 4. Separate — notice 1+3ev=ddv(3ev+v)1+3e^v=\dfrac{d}{dv}\left(3e^v+v\right). 1+3ev3ev+v dv=−dxx\dfrac{1+3e^v}{3e^v+v}\,dv=-\dfrac{dx}{x}. …

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