Q.In a bank, principal increases continuously at the rate of 5% per year. An amount of Rs. 1,000 is deposited with this bank. Find how much it will be worth after 10 years ($e^{0.5}=1.648$). OR Find the general solution of the differential equation $y\,dx-(x+2y^2)\,dy=0$.

Since the growth rate is proportional to the number present, $x$ satisfies $\dfrac{dx}{dt} = kx$, giving the exponential relation $x = 100000\,e^{kt}$.

The rate of growth of bacteria is proportional to the number of bacteria present at time $t$:

$\frac{dx}{dt} = kx$

Separating variables and integrating,

$\int \frac{dx}{x} = \int k\,dt \implies \ln x = kt + C$

…

Using $x = 100000e^{kt}$ with a 10% increase at $t=2$ gives $k = \tfrac12\ln(1.1) \approx 0.0477$ per hour.

From the previous part, $x = 100000\,e^{kt}$.

The bacteria increase by 10% in 2 hours, so at $t = 2$, $x = 100000 + 10\%\text{ of }100000 = 110000$.

$110000 = 100000\,e^{2k}$

$e^{2k} = 1.1$

$2k = \ln(1.1)$ …

Doubling time follows from $x=100000e^{kt}$: setting $x=200000$ gives $t = \dfrac{\ln 2}{k} = \dfrac{2\ln 2}{\ln 1.1} \approx 14.55$ hours.

We want the time $t$ at which $x = 200000$, starting from $x = 100000\,e^{kt}$:

$200000 = 100000\,e^{kt}$

$e^{kt} = 2$

$kt = \ln 2$

$t = \frac{\ln 2}{k}$

Using $k = \dfrac{1}{2}\ln(1.1)$ from the previous part,

$t = \frac{\ln 2}{\tfrac12 \ln(1.1)} = \frac{2\ln 2}{\ln(1.1)}$

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Geometric (exponential) growth with unlimited resources produces a J-shaped curve where population density accelerates upward without bound; the answer is (C).

The differential equation $f'(x)=f(x)$ is solved by separation of variables, giving the exponential function.

We have $\dfrac{df}{dx} = f$. Separating variables (assuming $f\ne 0$):

$\dfrac{df}{f} = dx$

Integrating both sides:

$\ln|f| = x + c_1$

$f(x) = Ce^{x}$, where $C = \pm e^{c_1}$ is an arbitrary constant.

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Populations can grow exponentially (J-curve, unlimited resources) or logistically (S-curve, limited resources/carrying capacity).

When resources (food, space) in a habitat are unlimited, a population grows exponentially, and a plot of population size against time gives a J-shaped curve (dN/dt = rN). In nature, resources are finite, so growth eventually slows as the population approaches the habitat's carrying capacity (K); a plot of this pattern gives an S-shaped (sigmoid) curve, described by the logistic growth equation …

The maximum population size that a given habitat can sustainably support is called the carrying capacity of that habitat for the species.

Every habitat has a finite amount of resources (food, space, nesting sites, etc.), so a population of a species cannot keep growing indefinitely within it. The largest population size that the habitat's resources can support — beyond which the population cannot be sustained (due to resource limitation, competition, increased mortality, etc.) — is termed the carrying capacity (K) of that habita …