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NCERT Exemplar · Q1

Q.Find the solution of dydx=2y−x\frac{dy}{dx}=2^{y-x}.

Yanam CbseShort· 3mImportance★★★★★
Appeared in past exams:MHT-CET 2021· Set pcm-2021-09-23-M· 2mexact
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✓ Free question

Separable: 2−y dy=2−x dx2^{-y}\,dy = 2^{-x}\,dx. Integrating gives 2−x−2−y=C2^{-x} - 2^{-y} = C (equivalently 2−y−2−x=C2^{-y} - 2^{-x} = C).

Write 2y−x=2y 2−x2^{y-x} = 2^{y}\,2^{-x}, so the equation separates:

dy2y=2−x dx⟹2−y dy=2−x dx.\frac{dy}{2^{y}} = 2^{-x}\,dx \quad\Longrightarrow\quad 2^{-y}\,dy = 2^{-x}\,dx.

Integrate both sides using ∫2−u du=−2−ulog⁡2\displaystyle\int 2^{-u}\,du = -\frac{2^{-u}}{\log 2}:

−2−ylog⁡2=−2−xlog⁡2+C.-\frac{2^{-y}}{\log 2} = -\frac{2^{-x}}{\log 2} + C.

Multiply through by −log⁡2-\log 2 and absorb constants:

2−y=2−x+C1⟹2−x−2−y=C.2^{-y} = 2^{-x} + C_1 \quad\Longrightarrow\quad 2^{-x} - 2^{-y} = C.

Verification: differentiating 2−x−2−y=C2^{-x} - 2^{-y} = C gives −2−xlog⁡2+2−ylog⁡2 dydx=0-2^{-x}\log 2 + 2^{-y}\log 2\,\dfrac{dy}{dx} = 0, i.e. dydx=2−x2−y=2 y−x\dfrac{dy}{dx} = \dfrac{2^{-x}}{2^{-y}} = 2^{\,y-x}, as required.

✓Final answer

The general solution is 2−x−2−y=C2^{-x} - 2^{-y} = C (equivalently 2−y−2−x=C2^{-y} - 2^{-x} = C), where CC is an arbitrary constant.

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