1.
Question 1. The differential equation \(\frac{dy}{dx}+x^3=0\) is solved by \(y=-\frac{x^4}{4}+C\).
Step 1. Rewrite as \(dy=-x^3dx\). Step 2. Integrate both sides. Step 3. \(y=-\frac{x^4}{4}+C\). Step 4. This matches the proposed solution. Answer: A
2.
Question 2. If \(\frac{dy}{dx}-x^2-1=0\) and \(y(0)=3\), then \(y=\frac{x^3}{3}+x+3\).
Step 1. Rewrite as \(dy=(x^2+1)dx\). Step 2. Integrate to obtain \(y=\frac{x^3}{3}+x+C\). Step 3. Use \(y(0)=3\) to get \(C=3\). Step 4. The proposed expression omits the \(\frac13\), so it is false. Answer: B
3.
Question 3. Solve \(y'=y^3\), \(y(1)=2\).
Step 1. Separate variables. Step 2. Integrate: \(-\frac1{2y^2}=x+C\). Step 3. Use \(y(1)=2\) to determine \(C=-\frac98\). Step 4. Solve for \(y\); only option D satisfies the IVP. Answer: D
4.
Question 4. A bacteria culture starts at 5000 and grows at 3% per day. How long until it reaches 10000?
Step 1. Use \(y=5000e^{0.03t}\). Step 2. Set \(10000=5000e^{0.03t}\). Step 3. Solve \(t=\ln2/0.03\approx23.1\). Answer: C
5.
Question 5. If \(\frac{dy}{dx}=-\frac{3x^2y}{x-1}\), find the slope at \((2,1)\).
Step 1. Substitute \(x=2,y=1\). Step 2. Compute \(-3(2)^2(1)/(2-1)=-12\). Answer: C
6.
Question 6. If \(\frac{dy}{dx}=4-xy\), through \((2,3)\), Euler with \(\Delta x=0.1\) gives \(y(2.1)=1.8\).
Step 1. Slope at \((2,3)\): \(4-2\cdot3=-2\). Step 2. Euler: \(y_1=3+(-2)(0.1)=2.8\). Step 3. 1.8 is incorrect. Answer: A
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