JEE Main 2012

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Animated Solution for Mathematics - Differential Equations: The population $p (t)$ at time $t$ of a certain mouse species satisfies the differential equation $\frac{d p ( t )}{d t} = 0.5 p (t) - 450$ . If $p (0) = 850$ , then the time at which the population becomes zero is :

Q: Can I use the integrating factor method instead of separation of variables?

Yes, you absolutely can! The equation $\frac{dp}{dt} - 0.5p = -450$ is a standard linear ODE of the form $\frac{dy}{dx} + Py = Q$. You could find the integrating factor $e^{\int -0.5 dt} = e^{-0.5t}$ and solve it that way. However, for this specific problem, the variable separable method is much faster and less prone to algebraic errors. Both methods will lead you to the same correct answer!

Select Answer:

Progressive Hints

Hint 1

Hint 1: The Spark. Visualize the differential equation

\frac{d p ( t )}{d t} = 0.5 p (t) - 450

. This is a first-order linear differential equation. Notice that the rate of change depends on the population itself. The critical value here is where the rate of change is zero, which happens when

0.5 p (t) - 450 = 0

. This value,

p = 900

, acts as a horizontal asymptote for the population curve.

Hint 2

Hint 2: The Arsenal. You need to use the method of separation of variables. Factor out

0.5

from the right side to make the expression

0.5 (p - 900)

. Then, move all terms involving

p

to one side and all terms involving

t

to the other. Remember that when you integrate

\int \frac{d p}{p - 900}

, you will get a natural logarithm involving an absolute value:

ln ∣ p - 900∣

Hint 3

Hint 3: The Execution. After integrating, you will have

ln ∣ p - 900∣ = 0.5 t + C

. Use the initial condition

p (0) = 850

to solve for the constant

C

. Once you have

C

, set

p (t) = 0

to find the time of extinction. The equation to solve will be

ln ∣0 - 900∣ = 0.5 t + C

Visualized Solution

Differential Equation

Given Differential Equation:
$\frac{d p ( t )}{d t} = 0.5 p (t) - 450$
Initial Condition:
At $t = 0$ , $p (0) = 850$

Variable Separable Method

Factor out $0.5$ from the right side:
$\frac{d p}{d t} = \frac{1}{2} (p - 900)$
Rearrange terms to separate variables:
$\frac{d p}{p - 900} = \frac{1}{2} d t$

Integrating Both Sides

Integrate both sides:
$\int \frac{d p}{p - 900} = \int \frac{1}{2} d t$
Result:
$ln ∣ p - 900∣ = \frac{t}{2} + C$

Initial Condition $p (0) = 850$

Use the given condition: at $t = 0$ , $p = 850$ .
Substitute these values into the integrated equation to find $C$ .

Finding Constant $C$

$ln ∣850 - 900∣ = \frac{0}{2} + C$
$ln ∣ - 50∣ = C$
$C = ln 50$

Specific Solution

Substitute $C$ back into the equation:
$ln ∣ p - 900∣ = \frac{t}{2} + ln 50$

Setting $p (t) = 0$

Set $p = 0$ to find the time of extinction:
$ln ∣0 - 900∣ = \frac{t}{2} + ln 50$
$ln 900 = \frac{t}{2} + ln 50$

Logarithmic Properties

Rearrange to isolate the $t$ term:
$ln 900 - ln 50 = \frac{t}{2}$
Apply log property $ln A - ln B = ln (\frac{A}{B})$ :
$ln (\frac{900}{50}) = \frac{t}{2}$

Final Time $t$

Simplify the fraction:
$ln 18 = \frac{t}{2}$
Solve for $t$ :
$t = 2 ln 18$

Summary and Insights

Key Takeaway: The population reaches zero at $t = 2 ln 18$ .
Insight: Since $p (0) < 900$ , the rate $\frac{d p}{d t}$ was negative, leading to extinction.
Challenge: What would happen if $p (0) > 900$ ? Try solving for that case!

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The Sigma Insight: Variable Separable Method

Solution Diagram

Analyzing the Setup

The population growth of the mice is governed by the differential equation:

\frac{d p ( t )}{d t} = 0.5 p (t) - 450

This equation dictates that the rate of change is proportional to the current population. We observe an equilibrium point at

p = 900

, where the rate of change becomes zero.

Since the initial population

p (0) = 850

is strictly less than the equilibrium threshold of

900

, the population is destined to decline. Our objective is to determine the exact time

t

when the population

p (t)

reaches zero.

The Art of Separation

To solve the differential equation, we first factor the right-hand side:

\frac{d p}{d t} = 0.5 (p - 900)

We apply the method of separation of variables to isolate

p

and

t

. By dividing both sides by

(p - 900)

and multiplying by

d t

, we obtain:

\frac{d p}{p - 900} = 0.5 d t

This transformation allows us to proceed with integration on both sides of the equation.

The Integration and the Constant

We integrate both sides of the separated equation:

\int \frac{d p}{p - 900} = \int 0.5 d t

Performing the integration yields:

ln ∣ p - 900∣ = 0.5 t + C

To find the constant

C

, we apply the initial condition

p (0) = 850

ln ∣850 - 900∣ = 0.5 (0) + C \Rightarrow C = ln 50

Substituting

C

back into our general solution, we obtain the specific equation for this population:

ln ∣ p - 900∣ = 0.5 t + ln 50

The Final Countdown to Extinction

To find the time of extinction, we set

p = 0

and solve for

t

ln ∣0 - 900∣ = 0.5 t + ln 50

ln 900 = 0.5 t + ln 50

Rearranging the terms to isolate

t

, we use the logarithmic property

ln A - ln B = ln (\frac{A}{B})

ln 900 - ln 50 = 0.5 t

ln (\frac{900}{50}) = 0.5 t

ln 18 = 0.5 t

Multiplying both sides by

2

, we find the final time of extinction:

$t = 2 ln 18$

Common Traps & FAQs

Wait, why do we have an absolute value inside the natural log? Is it just for show?

That is a fantastic question! The absolute value is absolutely essential. The natural logarithm function

ln (x)

is only defined for positive values of

x

. Since our population

p (t)

is less than

900

(starting at

850

), the expression

p - 900

is negative. Without the absolute value, we would be trying to take the logarithm of a negative number, which is undefined in the real number system. The absolute value ensures we are working with a positive magnitude, allowing us to proceed with the calculation.

Why is the population decreasing? The equation has a positive

0.5 p

term.

It is easy to get distracted by the positive coefficient of

p

. Look closely at the full expression:

\frac{d p}{d t} = 0.5 p - 450

. If the population

p

is less than

900

, then

0.5 p

is less than

450

. This means

0.5 p - 450

is negative, resulting in a negative rate of change. The population is shrinking because it is below the critical threshold of

900

Can I use the integrating factor method instead of separation of variables?

Yes, you absolutely can! The equation

\frac{d p}{d t} - 0.5 p = - 450

is a standard linear ODE of the form

\frac{d y}{d x} + P y = Q

. You could find the integrating factor

e^{\int - 0.5 d t} = e^{- 0.5 t}

and solve it that way. However, for this specific problem, the variable separable method is much faster and less prone to algebraic errors. Both methods will lead you to the same correct answer!

Animated Solution for Mathematics - Differential Equations: The population $p (t)$ at time $t$ of a certain mouse species satisfies the differential equation $\frac{d p ( t )}{d t} = 0.5 p (t) - 450$ . If $p (0) = 850$ , then the time at which the population becomes zero is :

Select Answer:

Visualized Solution

Differential Equation

Variable Separable Method

Integrating Both Sides

Initial Condition $p (0) = 850$

Finding Constant $C$

Specific Solution

Setting $p (t) = 0$

Logarithmic Properties

Final Time $t$

Summary and Insights

The Sigma Insight: Variable Separable Method

Analyzing the Setup

The Art of Separation

The Integration and the Constant

The Final Countdown to Extinction

Similar Questions

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Select Answer:

Visualized Solution

Differential Equation

Variable Separable Method

Integrating Both Sides

Initial Condition .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } p(0)=850

Finding Constant .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } C

Specific Solution

Setting .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } p(t)=0

Logarithmic Properties

Final Time .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } t

Summary and Insights

The Sigma Insight: Variable Separable Method

Analyzing the Setup

The Art of Separation

The Integration and the Constant

The Final Countdown to Extinction

Similar Questions

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Initial Condition $p (0) = 850$

Finding Constant $C$

Setting $p (t) = 0$

Final Time $t$

Let the population of rabbits surviving at time $t$ be governed by the differential equation $\frac{d p ( t )}{d t} = \frac{1}{2} p (t) - 200$ . If $p (0) = 100$ , then $p (t)$ equals:

If $y = y (x)$ is the solution of the differential equation, $e^{y} (\frac{d y}{d x} - 1) = e^{x}$ such that $y (0) = 0$ , then $y (1)$ is equal to

Let $x = x (y)$ be the solution of the differential equation $2 y e^{x / y^{2}} d x + (y^{2} - 4 x e^{x / y^{2}}) d y = 0$ such that $x (1) = 0$ . Then, $x (e)$ is equal to

If $y = y (x)$ satisfies the differential equation $8 x (9 + x) d y = (4 + 9 + x)^{- 1} d x$ , $x > 0$ and $y (0) = 7$ , then $y (256) =$

Let $y = y (x)$ be the solution of the differential equation $d y = e^{α x + y} d x; α \in N$ . If $y (lo g_{e} 2) = lo g_{e} 2$ and $y (0) = lo g_{e} (1/2)$ , then the value of $α$ is equal to