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Question 1. The differential equation \(\frac{dy}{dx}+x^3=0\) is solved by \(y=-\frac{1}{3x^2}+C\).
Step 1. Start with \(\frac{dy}{dx}+x^3=0\). This gives \(\frac{dy}{dx}=-x^3\).\nStep 2. Separate and integrate both sides: \(dy=-x^3 dx\), so \(\int dy=\int -x^3 dx\).\nStep 3. Therefore, \(y=-\frac{x^4}{4}+C\).\nStep 4. The proposed solution \(y=-\frac{1}{3x^2}+C\) does not match the correct general solution.\nAnswer: A. False
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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. Start with \(\frac{1}{x+\sin x}\frac{dy}{dx}=1\). Multiply both sides by \(x+\sin x\) to get \(\frac{dy}{dx}=x+\sin x\).\nStep 2. Integrate both sides: \(y=\int (x+\sin x)dx=\frac{x^2}{2}-\cos x+C\).\nStep 3. Use \(y(\pi)=0\): \(0=\frac{\pi^2}{2}-\cos(\pi)+C\). Since \(\cos(\pi)=-1\), we get \(0=\frac{\pi^2}{2}+1+C\).\nStep 4. Solve for \(C\): \(C=-1-\frac{\pi^2}{2}\).\nStep 5. The solution is \(y=\frac{x^2}{2}-\cos x-1-\frac{\pi^2}{2}\), not \(y=\cos x+x^2-\pi^2+1\).\nAnswer: B. False
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Question 3. Solve the differential equation \(y'=\frac{x}{y}\), \(y(0)=3\).
Step 1. Rewrite \(y'=\frac{x}{y}\) as \(\frac{dy}{dx}=\frac{x}{y}\).\nStep 2. Separate variables: \(y\,dy=x\,dx\).\nStep 3. Integrate both sides: \(\int y\,dy=\int x\,dx\), so \(\frac{y^2}{2}=\frac{x^2}{2}+C\).\nStep 4. Use \(y(0)=3\): \(\frac{3^2}{2}=\frac{0^2}{2}+C\), so \(C=\frac{9}{2}\).\nStep 5. Substitute \(C\): \(\frac{y^2}{2}=\frac{x^2}{2}+\frac{9}{2}\). Multiply by 2 to get \(y^2=x^2+9\).\nStep 6. Thus \(y=\pm\sqrt{x^2+9}\). Since \(y(0)=3\), choose the positive solution.\nAnswer: A. \(y=\sqrt{x^2+9}\)
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Question 4. The Half-Life of a 1000-gram substance is 50 years. If it decays according to an exponential growth model, the amount of time it will take for the substance to be only 100 g, is
Step 1. Use the exponential decay model \(A=1000e^{-kt}\).\nStep 2. The half-life is 50 years, so \(500=1000e^{-50k}\). Divide by 1000: \(\frac{1}{2}=e^{-50k}\).\nStep 3. Take natural logarithms: \(\ln\left(\frac{1}{2}\right)=-50k\). Therefore, \(k=-\frac{\ln\left(\frac{1}{2}\right)}{50}\approx 0.01386\).\nStep 4. Now find when the amount is 100 g: \(100=1000e^{-0.01386t}\). Divide by 1000: \(\frac{1}{10}=e^{-0.01386t}\).\nStep 5. Take natural logarithms: \(\ln\left(\frac{1}{10}\right)=-0.01386t\).\nStep 6. Solve for \(t\): \(t=\frac{-\ln\left(\frac{1}{10}\right)}{0.01386}\approx 166\).\nStep 7. Since \(166\) is between 165 and 170, the correct choice is A.\nAnswer: A. Between 165 and 170 years
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