CUET UG 2025 — Mathematics Calculus

Given differential equation: $(x + 1)\frac{dy}{dx} + 1 - 2e^{-y} = 0$(x+1)dxdy​+1−2e−y=0 with initial condition $y(0) = 0$y(0)=0

$$(x + 1)\frac{dy}{dx} = 2e^{-y} - 1$$(x+1)dxdy​=2e−y−1

$$\frac{dy}{2e^{-y} - 1} = \frac{dx}{x + 1}$$2e−y−1dy​=x+1dx​

Multiply numerator and denominator by $e^y$ey:

$$\frac{e^y \, dy}{2 - e^y} = \frac{dx}{x + 1}$$2−eyeydy​=x+1dx​

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Let $u = e^y$u=ey

Then $du = e^y \, dy$du=eydy

The equation becomes:

$$\frac{du}{2 - u} = \frac{dx}{x + 1}$$2−udu​=x+1dx​

$$-\frac{du}{u - 2} = \frac{dx}{x + 1}$$−u−2du​=x+1dx​

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Integrating both sides:

$$\int -\frac{du}{u - 2} = \int \frac{dx}{x + 1}$$∫−u−2du​=∫x+1dx​

$$-\ln|u - 2| = \ln|x + 1| + C$$−ln∣u−2∣=ln∣x+1∣+C

Substituting back $u = e^y$u=ey:

$$-\ln|e^y - 2| = \ln|x + 1| + C$$−ln∣ey−2∣=ln∣x+1∣+C

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Using the initial condition $y(0) = 0$y(0)=0:

$$-\ln|e^0 - 2| = \ln|0 + 1| + C$$−ln∣e0−2∣=ln∣0+1∣+C

$$-\ln|1 - 2| = \ln(1) + C$$−ln∣1−2∣=ln(1)+C

$$-\ln(1) = 0 + C$$−ln(1)=0+C

$$C = 0$$C=0

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With $C = 0$C=0:

$$-\ln|e^y - 2| = \ln|x + 1|$$−ln∣ey−2∣=ln∣x+1∣

$$\ln|e^y - 2|^{-1} = \ln|x + 1|$$ln∣ey−2∣−1=ln∣x+1∣

$$\ln\frac{1}{|e^y - 2|} = \ln|x + 1|$$ln∣ey−2∣1​=ln∣x+1∣

$$\frac{1}{|e^y - 2|} = |x + 1|$$∣ey−2∣1​=∣x+1∣

$$1 = |x + 1| \cdot |e^y - 2|$$1=∣x+1∣⋅∣ey−2∣

$$|(x + 1)(e^y - 2)| = 1$$∣(x+1)(ey−2)∣=1