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Exercise 9.3 · Q23

Q.The general solution of the differential equation dydx=ex+y\dfrac{dy}{dx} = e^{x+y} is (A) ex+e−y=Ce^x + e^{-y} = C (B) ex+ey=Ce^x + e^y = C (C) e−x+ey=Ce^{-x} + e^y = C (D) e−x+e−y=Ce^{-x} + e^{-y} = C

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Appeared in past exams:AP EAPCET 2021· Set eng-2021-08-24-FN· 1mexact
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The key idea is to separate the variables so that all yy terms are on one side and all xx terms on the other, then integrate. The general solution is ex+e−y=Ce^x + e^{-y} = C, which corresponds to option (A).

Why Separation of Variables works here

The given equation is dydx=ex+y\frac{dy}{dx} = e^{x+y}. The right-hand side is a product of a function of xx and a function of yy — because ex+y=ex⋅eye^{x+y} = e^x \cdot e^y. That’s the signal: whenever you can write dydx=f(x)⋅g(y)\frac{dy}{dx} = f(x) \cdot g(y), you can rearrange to get all yy on one side and all xx on the other, then integrate both sides. This is the method of separation of variables.

The trick is to treat dydx\frac{dy}{dx} as a fraction (which works in this context) and multiply both sides by dxdx and by whatever is needed to isolate yy and xx.


  1. Rewrite the equation using the exponent rule ex+y=ex⋅eye^{x+y} = e^x \cdot e^y, so

dydx=ex⋅ey\frac{dy}{dx} = e^x \cdot e^y

  1. Separate the variables Multiply both sides by dxdx and divide by eye^y:

dyey=ex dx\frac{dy}{e^y} = e^x \, dx

This is the separation step — every yy is on the left, every xx on the right.

  1. Integrate both sides

∫e−y dy=∫ex dx\int e^{-y} \, dy = \int e^x \, dx

The left integral is ∫e−ydy=−e−y+C1\int e^{-y} dy = -e^{-y} + C_1 (because the derivative of −e−y-e^{-y} is e−ye^{-y}).

The right integral is ∫exdx=ex+C2\int e^x dx = e^x + C_2.

So we have:

−e−y=ex+C-e^{-y} = e^x + C

where C=C2−C1C = C_2 - C_1 is a single arbitrary constant.

  1. Rearrange to match the given options Multiply both sides by −1-1: e−y=−ex−Ce^{-y} = -e^x - C …

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