CUET UG 2025 — Mathematics Calculus

Given: $\log_e\left(\frac{dy}{dx}\right) = ax + by$loge​(dxdy​)=ax+by

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Taking exponential (base $e$e) on both sides:

$$e^{\log_e\left(\frac{dy}{dx}\right)} = e^{ax + by}$$eloge​(dxdy​)=eax+by

$$\frac{dy}{dx} = e^{ax + by}$$dxdy​=eax+by

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Using $e^{A+B} = e^A \cdot e^B$eA+B=eA⋅eB:

$$\frac{dy}{dx} = e^{ax} \cdot e^{by}$$dxdy​=eax⋅eby

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Separating variables by dividing both sides by $e^{by}$eby:

$$\frac{1}{e^{by}} \cdot dy = e^{ax} \cdot dx$$eby1​⋅dy=eax⋅dx

$$e^{-by} \, dy = e^{ax} \, dx$$e−bydy=eaxdx

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Integrating both sides:

$$\int e^{-by} \, dy = \int e^{ax} \, dx$$∫e−bydy=∫eaxdx

Left side: $\displaystyle \int e^{-by} dy = \frac{e^{-by}}{-b} = -\frac{e^{-by}}{b}$∫e−bydy=−be−by​=−be−by​

Right side: $\displaystyle \int e^{ax} dx = \frac{e^{ax}}{a}$∫eaxdx=aeax​

Therefore:

$$-\frac{e^{-by}}{b} = \frac{e^{ax}}{a} + C_1$$−be−by​=aeax​+C1​

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Rearranging:

$$\frac{e^{ax}}{a} + \frac{e^{-by}}{b} + C_1 = 0$$aeax​+be−by​+C1​=0

Replacing $C_1$C1​ with $C$C:

$$\frac{e^{ax}}{a} + \frac{e^{-by}}{b} + C = 0$$aeax​+be−by​+C=0

where $C$C is the constant of integration.