Question

Let the population of a species of birds surviving at a time \(t\) be governed by the differential equation \[ \frac{dp}{dt}-p=-100 \] If \(p(0)=50\), then \(p(-\ln2)\) is equal to:

• A. \(90\)
• B. \(40\)
• C. \(75\)
• D. \(100\)

Hint:

For linear differential equations: \[ \frac{dy}{dx}+Py=Q \] always use: \[ I.F.=e^{\int Pdx} \] Then convert the left side into an exact derivative.

Answer 1

The Correct Option is C

Concept: The given differential equation is a first-order linear differential equation of the form: \[ \frac{dy}{dx}+Py=Q \] Such equations are solved using the Integrating Factor (I.F.) method.

Step 1: Writing the equation in standard form. Given: \[ \frac{dp}{dt}-p=-100 \] Comparing with: \[ \frac{dy}{dx}+Py=Q \] we get: \[ P=-1,\quad Q=-100 \]

Step 2: Finding the Integrating Factor. Integrating Factor: \[ I.F.=e^{\int Pdt} \] Therefore: \[ I.F.=e^{\int(-1)dt} \] \[ I.F.=e^{-t} \]

Step 3: Multiplying the equation by the Integrating Factor. Multiplying throughout by \(e^{-t}\): \[ e^{-t}\frac{dp}{dt}-pe^{-t}=-100e^{-t} \] Left side becomes: \[ \frac{d}{dt}(pe^{-t}) \] Thus: \[ \frac{d}{dt}(pe^{-t})=-100e^{-t} \]

Step 4: Integrating both sides. Integrating: \[ pe^{-t}=\int -100e^{-t}dt \] \[ pe^{-t}=100e^{-t}+C \] Multiplying by \(e^t\): \[ p=100+Ce^t \]

Step 5: Using the initial condition. Given: \[ p(0)=50 \] Substituting: \[ 50=100+Ce^0 \] \[ 50=100+C \] \[ C=-50 \] Hence: \[ p(t)=100-50e^t \]

Step 6: Finding \(p(-\ln2)\). Substitute \(t=-\ln2\): \[ p(-\ln2)=100-50e^{-\ln2} \] Using: \[ e^{-\ln2}=\frac12 \] we get: \[ p(-\ln2)=100-50\left(\frac12\right) \] \[ p(-\ln2)=100-25 \] \[ p(-\ln2)=75 \]