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Exercise 10.6 · Q4

Q.2xy dx+(x2+2y2)dy=02xy\,dx+\left(x^2+2y^2\right)dy=0

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Rewrite in homogeneous form, substitute y=vxy=vx, use partial fractions to separate, integrate, and simplify back to x,yx,y.

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

Step 2. Substitute y=vxy=vx. v+xdvdx=−2v1+2v2 ⟹ xdvdx=−2v−v(1+2v2)1+2v2=−v(3+2v2)1+2v2v+x\dfrac{dv}{dx}=\dfrac{-2v}{1+2v^2}\ \Longrightarrow\ x\dfrac{dv}{dx}=\dfrac{-2v-v(1+2v^2)}{1+2v^2}=\dfrac{-v(3+2v^2)}{1+2v^2}.

Step 3. Separate; partial fractions. 1+2v2v(3+2v2) dv=−dxx\dfrac{1+2v^2}{v(3+2v^2)}\,dv=-\dfrac{dx}{x}; writing 1+2v2v(3+2v2)=1/3v+(4/3)v3+2v2\dfrac{1+2v^2}{v(3+2v^2)}=\dfrac{1/3}{v}+\dfrac{(4/3)v}{3+2v^2}.

Step 4. Integrate. 13ln⁡∣v∣+13ln⁡(3+2v2)=−ln⁡∣x∣+C1 ⟹ ln⁡∣v(3+2v2)∣=−3ln⁡∣x∣+C2 ⟹ v(3+2v2)=Kx3\dfrac13\ln|v|+\dfrac13\ln\left(3+2v^2\right)=-\ln|x|+C_1\ \Longrightarrow\ \ln\left|v\left(3+2v^2\right)\right|=-3\ln|x|+C_2\ \Longrightarrow\ v\left(3+2v^2\right)=\dfrac{K}{x^3}. …

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