CUET UG 2025 — Mathematics Algebra

The differential equation is $\frac{dy}{dx} = e^{x+y}$dxdy​=ex+y

Using the exponential property $e^{x+y} = e^x \cdot e^y$ex+y=ex⋅ey:

$$\frac{dy}{dx} = e^x \cdot e^y$$dxdy​=ex⋅ey

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Dividing both sides by $e^y$ey:

$$\frac{1}{e^y} \cdot \frac{dy}{dx} = e^x$$ey1​⋅dxdy​=ex

$$e^{-y} \cdot \frac{dy}{dx} = e^x$$e−y⋅dxdy​=ex

$$e^{-y} \, dy = e^x \, dx$$e−ydy=exdx

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Integrating both sides:

$$\int e^{-y} \, dy = \int e^x \, dx$$∫e−ydy=∫exdx

$$-e^{-y} = e^x + C_1$$−e−y=ex+C1​

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Multiplying both sides by $-1$−1:

$$e^{-y} = -e^x - C_1$$e−y=−ex−C1​

$$e^x + e^{-y} = -C_1$$ex+e−y=−C1​

Since $-C_1$−C1​ is an arbitrary constant, it can be written as $C$C:

$$e^x + e^{-y} = C$$ex+e−y=C

Therefore, the general solution is $e^x + e^{-y} = C$ex+e−y=C