EBUS CAL12 Chapter 7 Differential Equations and More Riemann Sums Test-1

Step 1. Start with \(y=Ce^{-3t}-7\). Step 2. Differentiate: \(y'=-3Ce^{-3t}\). Step 3. Differentiate again: \(y''=9Ce^{-3t}\). Step 4. Substitute into the left side: \(y''+4y'=9Ce^{-3t}+4(-3Ce^{-3t})=-3Ce^{-3t}\). Step 5. Substitute into the right side: \(-3y-21=-3(Ce^{-3t}-7)-21=-3Ce^{-3t}+21-21=-3Ce^{-3t}\). Step 6. Since both sides are equal, the statement is true. Answer: B

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, use \(u=\sin x\), so \(du=\cos x\,dx\). Then \(y^2=e^{\sin x}+C\). Step 5. Use the initial condition \((0,0)\): \(0^2=e^{\sin 0}+C=1+C\), so \(C=-1\). Step 6. Thus \(y^2=e^{\sin x}-1\), so \(y(x)=\pm\sqrt{e^{\sin x}-1}\). Step 7. Evaluate at \(x=\pi\): \(y(\pi)=\pm\sqrt{e^{\sin \pi}-1}=\pm\sqrt{1-1}=0\). Answer: B

Step 1. Use the exponential decay model \(y=y_0e^{-kt}\). Step 2. After 96 days, \(y=0.214y_0\), so \(0.214=e^{-96k}\). Step 3. After an additional 96 days, the total time is \(192\) days. Step 4. Since \(192=2(96)\), the remaining fraction is \((0.214)^2\). Step 5. Calculate \((0.214)^2=0.045796\approx 0.046\). Step 6. Convert to a percent: \(0.046=4.6\%\). Answer: A

Step 1. The slope is \(\frac{dy}{dx}\) evaluated at the point \((0,\pi)\). Step 2. Substitute \(x=0\) and \(y=\pi\): \(\frac{dy}{dx}=\frac{e^{2(0)+1}}{1+\tan(\pi)}\). Step 3. Since \(\tan(\pi)=0\), the denominator is \(1+0=1\). Step 4. Therefore, the slope is \(\frac{e^1}{1}=e\). Answer: A

Step 1. Use Euler's Method: \(p_n=p_{n-1}+\Delta t\cdot F(t_{n-1},p_{n-1})\), where \(F(t,p)=0.5p(1-p)\). Step 2. Since \(\Delta t=0.5\) and we need 2 years, use \(4\) steps. Step 3. Start with \(p_0=0.7\). Step 4. \(p_1=0.7+0.5(0.5)(0.7)(1-0.7)=0.7525\). Step 5. \(p_2=0.7525+0.5(0.5)(0.7525)(1-0.7525)\approx 0.7991\). Step 6. \(p_3=0.7991+0.5(0.5)(0.7991)(1-0.7991)\approx 0.8392\). Step 7. \(p_4=0.8392+0.5(0.5)(0.8392)(1-0.8392)\approx 0.8729\). Step 8. Since \(p\) is measured in thousands, \(0.8729(1000)\approx 873\). Answer: A

Step 1. Let \(f(x)=4x-\cos x\). Step 2. Then \(f'(x)=4+\sin x\). Step 3. Use Newton's Method: \(x_n=x_{n-1}-\frac{f(x_{n-1})}{f'(x_{n-1})}\). Step 4. Starting with \(x_0=7\), compute \(x_1=7-\frac{4(7)-\cos(7)}{4+\sin(7)}\approx 1.1494\). Step 5. Compute \(x_2=1.1494-\frac{4(1.1494)-\cos(1.1494)}{4+\sin(1.1494)}\approx 0.2968\). Step 6. Compute \(x_3=0.2968-\frac{4(0.2968)-\cos(0.2968)}{4+\sin(0.2968)}\approx 0.2430\). Answer: C

Step 1. Use the linearization formula \(L(x)=f(a)+f'(a)(x-a)\), with \(a=1\). Step 2. Differentiate: \(f'(x)=-4\cos x-4e^{-4x}\). Step 3. The linear approximation at \(x=1\) is \(L(x)=f(1)+f'(1)(x-1)\). Step 4. Using the values from the calculation, \(L(x)\approx -2.23x-1.11\). Step 5. Evaluate the approximation: \(L(1.2)\approx -2.23(1.2)-1.11=-3.79\). Step 6. Evaluate the exact value: \(f(1.2)=-4\sin(1.2)+e^{-4(1.2)}\approx -3.72\). Step 7. Percent error is \(\frac{|\text{exact}-\text{approx}|}{|\text{exact}|}\times 100\). Thus \(\frac{|-3.72-(-3.79)|}{|-3.72|}\times 100\approx 2\%\). Step 8. Since \(2\%\) is between \(0\%\) and \(6\%\), the correct choice is A. Answer: A

Step 1. The step size is \(\Delta x=0.5\), with interval \([-1,2]\). Step 2. The trapezoidal rule is \(\int_a^b f(x)\,dx\approx \frac{\Delta x}{2}[f(x_0)+2f(x_1)+\cdots+2f(x_{n-1})+f(x_n)]\). Step 3. The points are \(-1,-0.5,0,0.5,1,1.5,2\). Step 4. Substitute \(f(x)=x^6\): \(\frac{0.5}{2}[1+2(0.015625)+2(0)+2(0.015625)+2(1)+2(11.390625)+64]\). Step 5. This gives approximately \(22.46\). Step 6. Therefore, the statement is true. Answer: A