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Exercise 10.6 · Q6

Q.xdydx=y−xcos⁡2 ⁣(yx)x\dfrac{dy}{dx}=y-x\cos^2\!\left(\dfrac{y}{x}\right)

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Rewrite in homogeneous form, substitute y=vxy=vx, separate, and integrate using ∫sec⁡2v dv=tan⁡v\int\sec^2v\,dv=\tan v.

Step 1. Rewrite. dydx=yx−cos⁡2 ⁣(yx)\dfrac{dy}{dx}=\dfrac{y}{x}-\cos^2\!\left(\dfrac{y}{x}\right) — homogeneous of degree 00.

Step 2. Substitute y=vxy=vx. v+xdvdx=v−cos⁡2v ⟹ xdvdx=−cos⁡2vv+x\dfrac{dv}{dx}=v-\cos^2v\ \Longrightarrow\ x\dfrac{dv}{dx}=-\cos^2v.

Step 3. Separate. sec⁡2v dv=−dxx\sec^2v\,dv=-\dfrac{dx}{x}.

Step 4. Integrate. tan⁡v=−ln⁡∣x∣+C\tan v=-\ln|x|+C. …

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