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Exercises · Q13

Q.Solve dydx=ex−y\displaystyle\frac{dy}{dx} = e^{x - y}.

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✓ Free question

Write ex−y=ex e−ye^{x-y} = e^{x}\,e^{-y}, so the equation becomes

dydx=ex e−y.\frac{dy}{dx} = e^{x}\,e^{-y}.

Separate the variables (multiply both sides by ey dxe^{y}\,dx):

ey dy=ex dx.e^{y}\,dy = e^{x}\,dx.

Integrate both sides:

∫ey dy=∫ex dx⟹ey=ex+c.\int e^{y}\,dy = \int e^{x}\,dx \quad\Longrightarrow\quad e^{y} = e^{x} + c.

Verify (differentiate): from ey=ex+ce^{y} = e^{x}+c, differentiate with respect to xx: eydydx=exe^{y}\dfrac{dy}{dx} = e^{x}, so dydx=ex e−y=ex−y\dfrac{dy}{dx} = e^{x}\,e^{-y} = e^{x-y} — the original equation. Correct.

✓Final answer

ey=ex+ce^{y} = e^{x} + c.

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