1.
Question 1. The function \(y=Ce^{3t}-e^2\) solves the differential equation \(y''-4y'=-3y-3e^2\).
Step 1. Start with \(y=Ce^{3t}-e^2\). Step 2. Differentiate once: \(y'=3Ce^{3t}\). Step 3. Differentiate again: \(y''=9Ce^{3t}\). Step 4. Substitute into the left side: \(y''-4y'=9Ce^{3t}-4(3Ce^{3t})=9Ce^{3t}-12Ce^{3t}=-3Ce^{3t}\). Step 5. Substitute into the right side: \(-3y-3e^2=-3(Ce^{3t}-e^2)-3e^2=-3Ce^{3t}+3e^2-3e^2=-3Ce^{3t}\). Step 6. The left side equals the right side, so the function solves the differential equation. Answer: B
2.
Question 2. Consider the differential equation \(\frac{1}{\cos x}\frac{dy}{dx}=\frac{e^{\sin x}}{2y}\) with initial condition \((0,0)\). First solve this to find \(y(x)\), and then use it to answer this question: What is \(y(\pi)\)?
Step 1. Rewrite the differential equation as \(\frac{dy}{dx}=\frac{e^{\sin x}\cos x}{2y}\). Step 2. Separate variables: \(2y\,dy=e^{\sin x}\cos x\,dx\). Step 3. Integrate both sides: \(\int 2y\,dy=\int e^{\sin x}\cos x\,dx\). Step 4. For the right side, let \(u=\sin x\), so \(du=\cos x\,dx\). Then \(\int e^{\sin x}\cos x\,dx=\int e^u\,du=e^u=e^{\sin x}\). Step 5. Therefore, \(y^2=e^{\sin x}+C\). Step 6. Use the initial condition \((0,0)\): \(0^2=e^{\sin 0}+C=1+C\), so \(C=-1\). Step 7. Thus \(y^2=e^{\sin x}-1\), so \(y(x)=\pm\sqrt{e^{\sin x}-1}\). Step 8. Evaluate at \(x=\pi\): \(y(\pi)=\pm\sqrt{e^{\sin \pi}-1}=\pm\sqrt{1-1}=0\). Answer: C
3.
Question 3. After 75 days, a radioactive substance has decayed to \(26.7\%\) of its original amount. After an additional 75 days, what percent of its original amount will it have decayed to?
Step 1. Use the exponential decay model \(y=y_0e^{-kt}\). Step 2. After 75 days, the amount is \(26.7\%\) of the original, so \(0.267y_0=y_0e^{-75k}\). Step 3. Divide by \(y_0\): \(0.267=e^{-75k}\). Step 4. After an additional 75 days, the total time is \(150\) days, which is twice 75 days. Step 5. Therefore, the remaining fraction is \((0.267)^2\). Step 6. Calculate \((0.267)^2=0.071289\approx 0.071\). Step 7. Convert to a percent: \(0.071\approx 7.1\%\). Answer: A
4.
Question 4. For the differential equation \(\frac{dy}{dx}=\sin(x)y^2\), what is the slope at \((0,2)\)?
Step 1. The slope at a point is the value of \(\frac{dy}{dx}\) at that point. Step 2. Substitute \(x=0\) and \(y=2\) into \(\frac{dy}{dx}=\sin(x)y^2\). Step 3. \(\frac{dy}{dx}=\sin(0)(2)^2\). Step 4. Since \(\sin(0)=0\), the slope is \(0\cdot 4=0\). Answer: D
5.
Question 5. A certain wolf population, \(p(t)\), is governed by the differential equation \(\frac{dp}{dt}=0.4p(1-p)\), where \(p\) is the population in thousands and \(t\) is the time in years. If the current size of the population is \(0.8\), that is \(p(0)=0.8\), use Euler's Method with \(\Delta t=0.5\) to predict the wolf population 2 years from now.
Step 1. Euler's Method uses \(p_n=p_{n-1}+\Delta t\cdot F(t_{n-1},p_{n-1})\), where \(F(t,p)=0.4p(1-p)\). Step 2. Since \(\Delta t=0.5\) and we want 2 years, we need \(4\) steps. Step 3. Start with \(p_0=0.8\). Step 4. \(p_1=0.8+0.5[0.4(0.8)(1-0.8)]=0.8+0.032=0.832\). Step 5. \(p_2=0.832+0.5[0.4(0.832)(1-0.832)]\approx 0.8600\). Step 6. \(p_3=0.8600+0.5[0.4(0.8600)(1-0.8600)]\approx 0.8841\). Step 7. \(p_4=0.8841+0.5[0.4(0.8841)(1-0.8841)]\approx 0.9046\). Step 8. Since \(p\) is measured in thousands, \(0.9046(1000)\approx 905\). Answer: B
6.
Question 6. Using Newton's Method to solve \(5x-\cos(x)=0\), with an initial approximation of \(x_0=-4\), the value of \(x_3\) to 4 decimal places is:
Step 1. Let \(f(x)=5x-\cos x\). Step 2. Differentiate: \(f'(x)=5+\sin x\). Step 3. Newton's Method is \(x_n=x_{n-1}-\frac{f(x_{n-1})}{f'(x_{n-1})}\). Step 4. Starting with \(x_0=-4\), compute \(x_1=-4-\frac{5(-4)-\cos(-4)}{5+\sin(-4)}\approx -0.6394\). Step 5. Compute \(x_2=-0.6394-\frac{5(-0.6394)-\cos(-0.6394)}{5+\sin(-0.6394)}\approx 0.2689\). Step 6. Compute \(x_3=0.2689-\frac{5(0.2689)-\cos(0.2689)}{5+\sin(0.2689)}\approx 0.1967\). Answer: B
7.
Question 7. If the local linear approximation of \(f(x)=3\sin x+e^{3x}\) at \(x=2\) is used to find the approximation for \(f(1.9)\), then the percent error of this approximation is:
Step 1. Use the linearization formula \(L(x)=f(a)+f'(a)(x-a)\), where \(a=2\). Step 2. Differentiate: \(f'(x)=3\cos x+3e^{3x}\). Step 3. The linear approximation at \(x=2\) is \(L(x)=f(2)+f'(2)(x-2)\). Step 4. Using the calculation from the solution, \(L(x)\approx 1209.04x-2011.92\). Step 5. Evaluate the approximation: \(L(1.9)=1209.04(1.9)-2011.92\approx 285.25\). Step 6. Evaluate the exact value: \(f(1.9)=3\sin(1.9)+e^{3(1.9)}\approx 301.71\). Step 7. Percent error is \(\frac{|\text{exact}-\text{approx}|}{|\text{exact}|}\times 100\). Step 8. \(\frac{|301.71-285.25|}{301.71}\times 100\approx 5.45\%\). Step 9. Since \(5.45\%\) is between \(5\%\) and \(10\%\), the correct choice is A. Answer: A
8.
Question 8. Using Simpson's Rule with a step size of \(0.5\), an approximation for the integral \(\int_1^4 x^7\,dx\approx 8210.08\).
Step 1. The interval is \([1,4]\), and the step size is \(\Delta x=0.5\). Step 2. The points are \(1,1.5,2,2.5,3,3.5,4\). Step 3. Use Simpson's Rule: \(\int_a^b f(x)\,dx\approx \frac{\Delta x}{3}[f(x_0)+4f(x_1)+2f(x_2)+4f(x_3)+2f(x_4)+4f(x_5)+f(x_6)]\). Step 4. Since \(\Delta x=0.5\), \(\frac{\Delta x}{3}=\frac{1}{6}\). Step 5. Substitute \(f(x)=x^7\): \(\frac{1}{6}[1+4(1.5^7)+2(2^7)+4(2.5^7)+2(3^7)+4(3.5^7)+4^7]\). Step 6. This gives approximately \(8210.08\). Step 7. Therefore, the statement is true. Answer: B
1 out of 1
