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Exercise 10.9 · Q16

Q.The solution of dydx=2y−x\dfrac{dy}{dx}=2^{y-x} is

(1) 2x+2y=C2^x+2^y=C
(2) 2x−2y=C2^x-2^y=C
(3) 12x−12y=C\dfrac{1}{2^x}-\dfrac{1}{2^y}=C
(4) x+y=Cx+y=C
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Separate the exponential equation and integrate each side using the general power-base antiderivative formula, then cancel the common ln⁡2\ln2.

Step 1. Separate. dydx=2y2x ⟹ 2−y dy=2−x dx\dfrac{dy}{dx}=\dfrac{2^y}{2^x}\ \Longrightarrow\ 2^{-y}\,dy=2^{-x}\,dx.

Step 2. Integrate both sides (using ∫a−u du=−a−uln⁡a\int a^{-u}\,du=-\dfrac{a^{-u}}{\ln a}). −2−yln⁡2=−2−xln⁡2+C1-\dfrac{2^{-y}}{\ln2}=-\dfrac{2^{-x}}{\ln2}+C_1. …

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