CUET UG 2025 — Mathematics Calculus

Given differential equation: $\frac{dr}{dt} = -rt$dtdr​=−rt with initial condition $r(0) = r_0$r(0)=r0​

Separating variables:

$$\frac{dr}{r} = -t \, dt$$rdr​=−tdt

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Integrating both sides:

$$\int \frac{dr}{r} = \int -t \, dt$$∫rdr​=∫−tdt

$$\ln|r| = -\frac{t^2}{2} + C$$ln∣r∣=−2t2​+C

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Applying the exponential function to both sides:

$$e^{\ln|r|} = e^{-\frac{t^2}{2} + C}$$eln∣r∣=e−2t2​+C

$$r = e^C \cdot e^{-\frac{t^2}{2}}$$r=eC⋅e−2t2​

Let $A = e^C$A=eC:

$$r = A e^{-\frac{t^2}{2}}$$r=Ae−2t2​

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Applying the initial condition $r(0) = r_0$r(0)=r0​:

$$r_0 = A e^{-\frac{0^2}{2}}$$r0​=Ae−202​

$$r_0 = A \cdot 1$$r0​=A⋅1

$$A = r_0$$A=r0​

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Therefore, the solution is:

$$r = r_0 e^{-t^2/2}$$r=r0​e−t2/2