1.
Question 1. \(y=\frac{1}{x+2}\) is a solution to the differential equation \(\frac{dy}{dx}+y^2=0\).
Step 1. Start with \(y=\frac{1}{x+2}\). Step 2. Differentiate: \(\frac{dy}{dx}=-\frac{1}{(x+2)^2}\). Step 3. Square the function: \(y^2=\left(\frac{1}{x+2}\right)^2=\frac{1}{(x+2)^2}\). Step 4. Substitute into the differential equation: \(\frac{dy}{dx}+y^2=-\frac{1}{(x+2)^2}+\frac{1}{(x+2)^2}=0\). Step 5. Since the equation is satisfied, the statement is true. Answer: B
2.
Question 2. If \(\frac{1}{x+\sin x}\frac{dy}{dx}=1\), and \(y(\pi)=0\), then \(y=\cos x+x^2-\pi^2+1\).
Step 1. Rewrite the equation: \(\frac{dy}{dx}=x+\sin x\). Step 2. Integrate both sides: \(\int dy=\int (x+\sin x)\,dx\). Step 3. This gives \(y=\frac{x^2}{2}-\cos x+C\). Step 4. Use the condition \(y(\pi)=0\): \(0=\frac{\pi^2}{2}-\cos(\pi)+C=\frac{\pi^2}{2}+1+C\). Step 5. Solve for \(C\): \(C=-1-\frac{\pi^2}{2}\). Step 6. Therefore, \(y=\frac{x^2}{2}-\cos x-1-\frac{\pi^2}{2}\), which is not the same as \(y=\cos x+x^2-\pi^2+1\). Answer: A
3.
Question 3. Solve the differential equation \(y'=y^3\), \(y(2)=\frac{1}{2}\).
Step 1. Separate variables: \(\frac{dy}{y^3}=dx\). Step 2. Integrate both sides: \(\int y^{-3}\,dy=\int dx\). Step 3. This gives \(-\frac{1}{2y^2}=x+C\). Step 4. Use \(y(2)=\frac{1}{2}\): \(-\frac{1}{2(\frac{1}{2})^2}=2+C\). Step 5. Since \(2(\frac{1}{2})^2=\frac{1}{2}\), we get \(-2=2+C\), so \(C=-4\). Step 6. Substitute back: \(-\frac{1}{2y^2}=x-4\). Step 7. Solve for \(y\): \(y^2=-\frac{1}{2x-8}\), so \(y=\pm\sqrt{-\frac{1}{2x-8}}\). Step 8. Checking the original equation shows only the appropriate branch satisfies the initial condition, and this result is not listed among A, B, or C. Answer: D
4.
Question 4. Find the exponential growth/decay model that satisfies \(\frac{dy}{dt}=ky\), decays at \(5\%\) per year, and has the initial condition \(y(0)=3\).
Step 1. For exponential growth or decay, use \(y=Ce^{kt}\). Step 2. Since the quantity decays at \(5\%\) per year, \(k=-0.05\). Step 3. Substitute this into the model: \(y=Ce^{-0.05t}\). Step 4. Use the initial condition \(y(0)=3\): \(3=Ce^{-0.05(0)}=C\). Step 5. Therefore, the model is \(y=3e^{-0.05t}\). Answer: A
5.
Question 5. If \(\frac{dy}{dx}=-\frac{xy^2}{e^{3-x}}\), what is the value of the slope field at the point \((3,-1)\)?
Step 1. The value of the slope field is the value of \(\frac{dy}{dx}\) at the given point. Step 2. Substitute \(x=3\) and \(y=-1\) into \(\frac{dy}{dx}=-\frac{xy^2}{e^{3-x}}\). Step 3. Compute: \(\frac{dy}{dx}=-\frac{3(-1)^2}{e^{3-3}}\). Step 4. Since \((-1)^2=1\) and \(e^0=1\), we get \(\frac{dy}{dx}=-\frac{3}{1}=-3\). Answer: A
6.
Question 6. If \(\frac{dy}{dx}=3x^2y\), and it goes through the point \((1,2)\), if you tried to estimate the y-value at \(x=1.2\) using Euler's Method, where \(\Delta x=0.2\), you would get a y-value of \(2.2\).
Step 1. Euler's Method uses \(y_1=y_0+\frac{dy}{dx}\Delta x\). Step 2. At the point \((1,2)\), compute the slope: \(\frac{dy}{dx}=3(1)^2(2)=6\). Step 3. Use \(\Delta x=0.2\): \(y(1.2)\approx 2+6(0.2)\). Step 4. This gives \(y(1.2)\approx 2+1.2=3.2\). Step 5. Since the statement says the value would be \(2.2\), the statement is false. Answer: A
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