The solution of the differential equation dr/dt = -rt, r(0) = r_0 is Option 1: r = r_0 e^(t^2/2) Option 2: r = -r_0 e^(t^2/2) Option 3: r = r_0 e^(-t^2/2) Option 4: r = -r_0 e^(-t^2/2)

Answer: r = r_0 e^(-t^2/2). Solution: Given differential equation: dr/dt = -rt with initial condition r(0) = r_0 Separating variables: dr/r = -t dt Integrating both sides: dr/r = -t dt, r, = -t^2/2 + C Applying the exponential function to both sides: e^(, r, ) = e^(-t^2/2 + C) r = e^C * e^(-t^2/2) Let A = e^C: r = A e^(-t^2/2) Applying the initial condition r(0) = r_0: r_0 = A e^(-0^2/2) r_0 = A * 1 A = r_0 Therefore, the solution is: r = r_0 e^(-t^2/2)

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The solution of the differential equation $\frac{dr}{dt} = -rt, r(0) = r_0$ is

$r = r_0 e^{t^2/2}$

$r = -r_0 e^{t^2/2}$

$r = r_0 e^{-t^2/2}$

$r = -r_0 e^{-t^2/2}$

✅ Correct Option: 3

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

Separating variables:

$\frac{dr}{r} = -t \, dt$

Integrating both sides:

$\int \frac{dr}{r} = \int -t \, dt$

$\ln|r| = -\frac{t^2}{2} + C$

Applying the exponential function to both sides:

$e^{\ln|r|} = e^{-\frac{t^2}{2} + C}$

$r = e^C \cdot e^{-\frac{t^2}{2}}$

Let $A = e^C$ :

$r = A e^{-\frac{t^2}{2}}$

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

$r_0 = A e^{-\frac{0^2}{2}}$

$r_0 = A \cdot 1$

$A = r_0$

Therefore, the solution is:

$r = r_0 e^{-t^2/2}$

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(A) it is a straight line.

(B) it is a parabola.

(D) Its equation is $xy(x + 1)\left(y - \frac{1}{\sqrt{e}} + 1\right) = 0$

Choose the correct answer from the options given below:

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(A) of order 1

(B) of degree 1

(D) of degree 3

Choose the correct answer from the options given below :

The solution of the differential equation drdt=−rt,r(0)=r0\frac{dr}{dt} = -rt, r(0) = r_0dtdr​=−rt,r(0)=r0​ is

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The solution of the differential equation $\frac{dr}{dt} = -rt, r(0) = r_0$ is