Unit 2 Practice / Review Problems — MATH 2410

Solution:

Since the growth rate is proportional to the population, the model is \[ \frac{dP}{dt} = kP \] The general solution is \[ P(t) = Ce^{kt}. \] Use the given data. \[ P(3) = 400 \quad \Rightarrow \quad 400 = Ce^{3k} \] \[ P(10) = 2000 \quad \Rightarrow \quad 2000 = Ce^{10k} \] Divide the equations: \[ \frac{2000}{400} = \frac{Ce^{10k}}{Ce^{3k}} = e^{7k} \] \[ 5 = e^{7k} \] Solve for \(k\): \[ k = \frac{\ln 5}{7} \] Substitute into \(400 = Ce^{3k}\): \[ C = 400e^{-3k} = 400e^{-3\ln(5)/7}. \] Therefore the initial population is \[ P(0) = C = 400e^{-3\ln(5)/7}. \] Numerically, \[ P(0) \approx 200 \text{ bacteria}. \]

Solution:

By Newton's Law of Cooling, \[ \frac{dT}{dt} = k(T - T_a) \] where \(T_a = 0\). The solution is \[ T(t) = Ce^{kt}. \] Use the data. \[ T(1) = 20 \quad \Rightarrow \quad 20 = Ce^k \] \[ T(5) = 5 \quad \Rightarrow \quad 5 = Ce^{5k} \] Divide the equations: \[ \frac{5}{20} = e^{4k} \] \[ \frac{1}{4} = e^{4k} \] \[ k = \frac{\ln(1/4)}{4}. \] Substitute into \(20 = Ce^k\): \[ C = 20e^{-k}. \] Therefore the initial temperature is \[ T(0) = C = 20e^{-k}. \] Numerically, \[ T(0) \approx 28.3^\circ C. \]

Solution:

The logistic model is \[ \frac{dN}{dt} = rN\left(1-\frac{N}{K}\right) \] where \(K = 50000\). The general solution is \[ N(t) = \frac{K}{1+Ae^{-rt}}. \] Use \(N(0)=5000\): \[ 5000 = \frac{50000}{1+A} \] \[ 1+A = 10 \] \[ A = 9. \] So \[ N(t) = \frac{50000}{1+9e^{-rt}}. \] Use \(N(1)=8000\): \[ 8000 = \frac{50000}{1+9e^{-r}}. \] Solve: \[ 1+9e^{-r} = 6.25 \] \[ 9e^{-r} = 5.25 \] \[ e^{-r} = \frac{5.25}{9}. \] Therefore \[ r = -\ln\!\left(\frac{5.25}{9}\right). \] The model is \[ N(t)=\frac{50000}{1+9e^{-rt}}, \quad r=-\ln\!\left(\frac{5.25}{9}\right). \]

Solution:

Let \[ x(t) = \text{pounds of salt in tank A}, \qquad y(t) = \text{pounds of salt in tank B}. \] Since the flow rates differ, the volumes in the tanks change over time. For tank A: \[ V_A(t) = 100 - 3t + 2t = 100 - t. \] For tank B: \[ V_B(t) = 100 + 3t - 2t = 100 + t. \] The concentrations in the tanks are \[ \frac{x(t)}{100-t}, \qquad \frac{y(t)}{100+t}. \] Now compute the rates of change. For tank A, salt enters from tank B at 2 gal/min: \[ 2 \cdot \frac{y}{100+t}. \] Salt leaves tank A at 3 gal/min: \[ 3 \cdot \frac{x}{100-t}. \] Therefore \[ \frac{dx}{dt} = 2\frac{y}{100+t} - 3\frac{x}{100-t}. \] For tank B, salt enters from tank A at 3 gal/min: \[ 3 \cdot \frac{x}{100-t}. \] Salt leaves tank B at 2 gal/min: \[ 2 \cdot \frac{y}{100+t}. \] Thus \[ \frac{dy}{dt} = 3\frac{x}{100-t} - 2\frac{y}{100+t}. \] The mathematical model is \[ \frac{dx}{dt} = 2\frac{y}{100+t} - 3\frac{x}{100-t}, \] \[ \frac{dy}{dt} = 3\frac{x}{100-t} - 2\frac{y}{100+t}, \] with initial conditions \[ x(0)=100, \qquad y(0)=50. \]

Solution:

Note the trigonometric identity \[ \sin^2(x)+\cos^2(x)=1. \] Multiplying both sides by 3 gives \[ 3 = 3\sin^2(x)+3\cos^2(x). \] Rewriting, \[ 3 - 3\sin^2(x) - 3\cos^2(x) = 0. \] Thus a nontrivial linear combination of the functions equals zero, so the functions are \[ \textbf{linearly dependent}. \]

Solution:

No. Some functions are scalar multiples of their derivatives. For example, let \[ f(x)=e^x. \] Then \[ f'(x)=e^x. \] Thus \[ f'(x)-f(x)=0. \] Since a nontrivial linear combination equals zero, the functions are linearly dependent. Therefore, functions are not necessarily linearly independent from their derivatives.

Solution:

Compute derivatives: \[ y=\sin(2x) \] \[ y' = 2\cos(2x) \] \[ y'' = -4\sin(2x). \] Thus \[ y'' +4y = -4\sin(2x)+4\sin(2x)=0. \] Therefore a second-order homogeneous differential equation with this solution is \[ y''+4y=0. \]

Solution:

Let \[ y_2 = v(x)y_1. \] Since \(y_1=e^{-2x}\), reduction of order gives \[ y_2 = xe^{-2x}. \] Thus the general solution is \[ y=C_1e^{-2x}+C_2xe^{-2x}. \]

Solution:

Step 1: Put the equation in standard form. Divide by \(x^2\): \[ y''+\frac{1}{x}y'=0. \] So \[ P(x)=\frac{1}{x}. \] --- Step 2: Reduction of Order Formula If one solution is \(y_1\), a second solution is \[ y_2=y_1\int \frac{e^{-\int P(x)\,dx}}{y_1^2}\,dx. \] --- Step 3: Compute needed pieces \[ \int P(x)dx=\int \frac{1}{x}dx=\ln x. \] \[ e^{-\int P(x)dx}=e^{-\ln x}=\frac{1}{x}. \] Since \(y_1=\ln x\), \[ y_1^2=(\ln x)^2. \] Thus \[ y_2=\ln x\int \frac{1}{x(\ln x)^2}dx. \] --- Step 4: Evaluate the integral Let \[ u=\ln x \] Then \[ du=\frac{1}{x}dx. \] The integral becomes \[ \int \frac{1}{u^2}du \] \[ =-\frac{1}{u}. \] Substitute back: \[ -\frac{1}{\ln x}. \] Thus \[ y_2=\ln x\left(-\frac{1}{\ln x}\right)=-1. \] Since constants do not affect linear independence, we take \[ y_2=1. \] --- Final Answer The general solution is \[ y=c_1\ln x+c_2. \]

Solution:

First solve the homogeneous equation. The characteristic equation is \[ r^2-3r+2=0. \] \[ (r-1)(r-2)=0 \] Thus \[ y_h=C_1e^x+C_2e^{2x}. \] Now find a particular solution. Try \[ y_p=Ae^{3x}. \] Substitute: \[ y_p'=3Ae^{3x}, \quad y_p''=9Ae^{3x}. \] \[ 9Ae^{3x}-9Ae^{3x}+2Ae^{3x}=5e^{3x}. \] \[ 2A=5 \] \[ A=\frac{5}{2}. \] Thus \[ y_p=\frac{5}{2}e^{3x}. \] The general solution is \[ y=C_1e^x+C_2e^{2x}+\frac{5}{2}e^{3x}. \]

Solution:

Characteristic equation: \[ r^2-36=0 \] \[ r=\pm6. \] Thus the general solution is \[ y=C_1e^{6x}+C_2e^{-6x}. \]

Solution:

Characteristic equation: \[ r^2+9=0. \] Since this equation has no real roots, the general solution contains trigonometric functions. We get \[ y=C_1\cos(3x)+C_2\sin(3x). \]

Solution:

Characteristic equation: \[ r^2+12r+36=0. \] \[ (r+6)^2=0. \] Thus \(r=-6\) is a repeated root. The general solution is \[ y=C_1e^{-6x}+C_2xe^{-6x}. \]

Solution:

First solve the homogeneous equation. \[ r^2+r-6=0 \] \[ (r+3)(r-2)=0 \] Thus \[ y_h=C_1e^{-3x}+C_2e^{2x}. \] For the particular solution try \[ y_p=Ax+B. \] Substitute: \[ y_p'=A, \quad y_p''=0. \] \[ A-6(Ax+B)=2x. \] Match coefficients: \[ -6A=2 \Rightarrow A=-\frac13 \] \[ A-6B=0. \] \[ -\frac13-6B=0 \] \[ B=-\frac1{18}. \] Thus \[ y_p=-\frac13x-\frac1{18}. \] The general solution is \[ y=C_1e^{-3x}+C_2e^{2x}-\frac13x-\frac1{18}. \]

Solution:

First solve the homogeneous equation. \[ r^2-16=0 \] \[ r=\pm4. \] \[ y_h=C_1e^{4x}+C_2e^{-4x}. \] For the particular solution try \[ y_p=Ae^{3x}. \] Substitute: \[ y_p'=3Ae^{3x}, \quad y_p''=9Ae^{3x}. \] \[ 9Ae^{3x}-16Ae^{3x}=2e^{3x}. \] \[ -7A=2. \] \[ A=-\frac{2}{7}. \] Thus \[ y_p=-\frac{2}{7}e^{3x}. \] The general solution is \[ y=C_1e^{4x}+C_2e^{-4x}-\frac{2}{7}e^{3x}. \]