Sample Questions on ODE

Example 1: The population of a certain species grows at a rate proportional to the current population size. If the population doubles in 10 years, and the initial population is 1000, find the population as a function of time. Then, determine how long it will take for the population to reach 5000.

Using basic differential equation for exponential growth:

dP/dt = kP

Where:

• ( P ) represents the population size.
• ( k ) is the growth rate constant.

Given that the population doubles in 10 years, we can find ( k ) using this information. When the population doubles, it means that ( P(t) = 2P(0) ), where (P(0)) is the initial population. Using the exponential growth equation:

2P(0) = P(0)e^10k

2 = e^10k

Taking the natural logarithm of both sides:

ln(2) = 10k

k = ln(2)/10

Now that we have the growth rate constant ( k ), we can find the population as a function of time (P(t)). The general solution to the differential equation is:

P(t) = P(0)e^kt

Given that (P(0) = 1000), we have:

P(t) = 1000[Tex]e^{\frac{\ln(2)}{10}t}[/Tex]

To find out how long it takes for the population to reach 5000, set P(t) = 5000 and solve for ( t ):

5000 = 1000[Tex]e^{\frac{\ln(2)}{10}t}[/Tex]

5 = [Tex]e^{\frac{\ln(2)}{10}t}[/Tex]

ln(5) = [Tex]\frac{\ln(2)}{10}t[/Tex]

t = 10ln(5) / ln(2)

t = [Tex]\frac{10\ln(5)}{\ln(2)}[/Tex]

t = [Tex]\frac{10 \times \ln(5)}{\ln(2)}[/Tex]

t ≈ [Tex]\frac{10 \times 1.609}{0.693}[/Tex]

t ≈ 16.09 / 0.693

t ≈ 23.22

So, it takes approximately 23.22 years for the population to reach 5000.

Example 2: A mass attached to a spring obeys the equation ([Tex]m\frac{d^2x}{dt^2}[/Tex] + kx = 0), where (m) is the mass, (k) is the spring constant, and (x) is the displacement from the equilibrium position. If (m = 2) kg, (k = 5) N/m, and the initial displacement is (x(0) = 3) m, and initial velocity is (v(0) = -2) m/s, find the position function (x(t)) and describe the motion of the mass.

Given the equation 2d²x/dt² + 5x = 0), the general solution is:

x(t) = A cos(ωt) + B sin(ω t)

Differentiating (x(t)) twice, we get:

dx/dt = -Aω sin(ω t) + Bω cos(ω t)

d²x/dt² = -Aω² cos(ω t) – Bω² sin(ω t)

Substituting these into the differential equation, we get:

2(-Aω² cos(ω t) – Bω² sin(ω t)) + 5(A cos(ω t) + B sin(ω t)) = 0

Solving for (A), (B), and (ω) using the initial conditions (x(0) = 3) and (v(0) = -2), we find:

A = 3

B = -2

ω = [Tex]\sqrt{\frac{5}{2}}[/Tex]

Thus, the solution to the differential equation with the given initial conditions is: [Tex]x(t) = 3\cos\left(\sqrt{\frac{5}{2}}t\right) – 2\sin\left(\sqrt{\frac{5}{2}}t\right)[/Tex]

Example 3: The temperature (T) of a cup of coffee in a room obeys Newton’s Law of Cooling: ([Tex]\frac{dT}{dt} = -k(T – T_a)[/Tex]), where (k) is a constant, (T) is the temperature of the coffee, and (T_a) is the ambient temperature. If (T(0) = 90° C), (T_a = 20° C), and (k = 0.05), find the temperature of the coffee after 10 minutes and determine how long it will take for the coffee to cool to (40° C).

Given that (T(0) = 90° C), (T_a = 20° C), and (k = 0.05), let’s solve for (T(t)).

Separating variables and integrating, we get:

dT/[T – T_a] = -kdt

Integrating both sides:

[Tex]\int \frac{1}{T – T_a} dT = -\int k dt[/Tex]

ln|T – T_a| = -kt + C₁

T – T_a = Ce^-kt

Given the initial condition (T(0) = 90° C), we can solve for (C):

90 – 20 = C ⇒ C = 70

So, the particular solution is:

T(t) = 70e^-0.05t + 20

Now, to find the time it takes for the coffee to cool to (40° C), we set (T(t) = 40) and solve for (t):

40 = 70e^-0.05t + 20

20 = 70e^-0.05t

e^-0.05t

= 20/70

= 2/7

[Tex]t = -\frac{\ln{\frac{2}{7}}}{0.05} \approx 15.98[/Tex]

So, it takes approximately (15.98) minutes for the coffee to cool to (40° C).

Ordinary Differential Equations(ODE) is the mathematical equation that describe how a function’s rate of change relates to its current state. It involves a single independent variable and its derivatives.

Let’s know more about Ordinary Differential Equations, it’s types, order and degree of Ordinary differential equation in detail below.