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Exercise 10.6 · Q5

Q.(y2−2xy)dx=(x2−2xy)dy\left(y^2-2xy\right)dx=\left(x^2-2xy\right)dy

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Rewrite in homogeneous form, substitute y=vxy=vx, separate via partial fractions, integrate, simplify.

Step 1. Rewrite. dydx=y2−2xyx2−2xy\dfrac{dy}{dx}=\dfrac{y^2-2xy}{x^2-2xy} — homogeneous of degree 00.

Step 2. Substitute y=vxy=vx. v+xdvdx=v2−2v1−2v ⟹ xdvdx=v2−2v−v(1−2v)1−2v=3v2−3v1−2v=3v(v−1)1−2vv+x\dfrac{dv}{dx}=\dfrac{v^2-2v}{1-2v}\ \Longrightarrow\ x\dfrac{dv}{dx}=\dfrac{v^2-2v-v(1-2v)}{1-2v}=\dfrac{3v^2-3v}{1-2v}=\dfrac{3v(v-1)}{1-2v}.

Step 3. Separate; partial fractions on 1−2vv(v−1)\dfrac{1-2v}{v(v-1)}. 1−2vv(v−1)=−1v−1v−1\dfrac{1-2v}{v(v-1)}=-\dfrac1v-\dfrac1{v-1}, so (−1v−1v−1)dv=3dxx\left(-\dfrac1v-\dfrac1{v-1}\right)dv=3\dfrac{dx}{x}.

Step 4. Integrate. −ln⁡∣v∣−ln⁡∣v−1∣=3ln⁡∣x∣+C1 ⟹ ln⁡∣v(v−1)∣=−3ln⁡∣x∣+C2 ⟹ v(v−1)=Kx3-\ln|v|-\ln|v-1|=3\ln|x|+C_1\ \Longrightarrow\ \ln\left|v(v-1)\right|=-3\ln|x|+C_2\ \Longrightarrow\ v(v-1)=\dfrac{K}{x^3}. …

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