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Question 1. A radioactive element has a half-life of 50 days. What percentage of the original sample is left after 85 days?
Use the half-life model \(\displaystyle N=N_0\left(\frac12\right)^{t/T_{1/2}}\). Here \(t=85\) and \(T_{1/2}=50\), so \(\displaystyle \frac{N}{N_0}=\left(\frac12\right)^{85/50}=\left(\frac12\right)^{1.7}\approx0.3078\). Therefore approximately \(30.78\%\) of the sample remains. Answer: D.
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Question 2. Solve \(yy'=x^3e^{x^2}\), for \(x\ge1\), with \(y(1)=0\).
Separate variables: \(y\,dy=x^3e^{x^2}dx\). The left side gives \(\frac12y^2\). For the right side, let \(u=x^2\), so \(du=2x\,dx\) and \(\displaystyle \int x^3e^{x^2}dx=\frac12\int ue^u\,du=\frac12e^u(u-1)+C\). Thus \(y^2=e^{x^2}(x^2-1)+C\). Using \(y(1)=0\) gives \(C=0\). Since the required branch is nonnegative, \(\displaystyle y=\sqrt{e^{x^2}(x^2-1)}\). Answer: A.
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Question 3. Find the solution of the differential equation \(xy'=y-2\), given that the point \((3,8)\) is on the curve.
Rewrite the equation as \(\displaystyle \frac{dy}{y-2}=\frac{dx}{x}\). Integrating gives \(\ln|y-2|=\ln|x|+C\), so \(y-2=kx\). Hence \(y=kx+2\). Using \((3,8)\): \(8=3k+2\), so \(k=2\). Therefore \(y=2x+2\). Answer: B.
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Question 4. The balance in an account triples in 20 years. Assuming that interest is compounded continuously, what is the annual percentage rate?
For continuous compounding, \(A=Pe^{rt}\). Tripling in 20 years means \(3P=Pe^{20r}\), so \(3=e^{20r}\). Taking logarithms, \(\displaystyle r=\frac{\ln3}{20}\approx0.05493\). As a percentage, the annual rate is approximately \(5.49\%\). Answer: C.
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Question 5. Find the equation of the curve satisfying \(\displaystyle \frac{dy}{dx}=12x^2+4x\) that passes through \((-1,-5)\).
Integrate both sides: \(\displaystyle y=\int(12x^2+4x)\,dx=4x^3+2x^2+C\). Use the point \((-1,-5)\): \(-5=4(-1)^3+2(-1)^2+C=-4+2+C\). Thus \(C=-3\), so \(y=4x^3+2x^2-3\). Answer: D.
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Question 6. Given \(\displaystyle \frac{dy}{dx}=e^{3x}\), find the equation of the curve that passes through \((\ln1,0)\).
Integrate: \(\displaystyle y=\int e^{3x}dx=\frac13e^{3x}+C\). Since \(\ln1=0\), the curve passes through \((0,0)\). Therefore \(0=\frac13e^0+C=\frac13+C\), so \(C=-\frac13\). Hence \(\displaystyle f(x)=\frac13e^{3x}-\frac13\). Answer: B.
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Question 7. Write an equation for the amount \(Q\) of a radioactive substance with a half-life of 30 days, if 10 grams are present when \(t=0\).
Use \(Q(t)=Q_0e^{-kt}\). For a half-life of 30 days, \(\displaystyle \frac12=e^{-30k}\), so \(\displaystyle k=\frac{\ln2}{30}\approx0.0231\). With \(Q_0=10\), \(\displaystyle Q(t)=10e^{-0.0231t}\). Answer: A.
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Question 8. Given \(\displaystyle \frac{dy}{dx}=\frac{3x^2}{x^3+5}\), find the equation of the curve that passes through \((0,0)\).
Let \(u=x^3+5\), so \(du=3x^2dx\). Then \(\displaystyle y=\int\frac{3x^2}{x^3+5}dx=\ln|x^3+5|+C\). Using \((0,0)\): \(0=\ln5+C\), so \(C=-\ln5\). Therefore \(f(x)=\ln(x^3+5)-\ln5\). Answer: E.
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Question 9. Solve \(y'e^x-y'=ye^x\), given \(y(3)=2\).
Factor \(y'\): \(y'(e^x-1)=ye^x\). Separate variables: \(\displaystyle \frac1y\,dy=\frac{e^x}{e^x-1}\,dx\). Let \(u=e^x-1\), so \(du=e^xdx\). Integrating gives \(\ln|y|=\ln|e^x-1|+C\), so \(y=k(e^x-1)\). Use \(y(3)=2\): \(\displaystyle k=\frac2{e^3-1}\). Thus \(\displaystyle y=\frac{2(e^x-1)}{e^3-1}\). Answer: C.
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Question 10. Solve the differential equation \(\displaystyle \frac{dy}{dx}-5=x^2\), given \(y(0)=-1\).
Rewrite the equation as \(\displaystyle \frac{dy}{dx}=x^2+5\). Integrating gives \(\displaystyle y=\frac13x^3+5x+C\). Use \(y(0)=-1\): \(C=-1\). Therefore \(\displaystyle y=\frac13x^3+5x-1\). Answer: B.
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Question 11. Solve the differential equation \(\displaystyle \frac{dy}{y^2}=x\,dx\).
Integrate both sides: \(\displaystyle \int y^{-2}dy=\int x\,dx\). This gives \(-\frac1y=\frac12x^2+C_1\). Multiply by \(-2\): \(\displaystyle \frac2y=-x^2+C\). Rearranging, \(\displaystyle x^2+\frac2y=C\). Answer: E.
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Question 12. Find the constant \(k\) so that the exponential function \(y=3e^{kx}\) passes through the point \((3,5)\) shown on the graph.
Substitute the point \((3,5)\) into \(y=3e^{kx}\): \(5=3e^{3k}\). Thus \(\displaystyle \frac53=e^{3k}\). Taking natural logarithms gives \(\displaystyle \ln\left(\frac53\right)=3k\), so \(\displaystyle k=\frac13\ln\left(\frac53\right)\). Answer: A.
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Question 13. A mold culture doubles its mass every five days. Find the growth model for a plate seeded with \(0.3\) grams of mold.
Use the exponential growth model \(y=y_0e^{kt}\). Doubling in five days means \(2=e^{5k}\), so \(\displaystyle k=\frac{\ln2}{5}\approx0.13863\). With initial mass \(y_0=0.3\), the model is \(\displaystyle y=0.3e^{0.13863t}\). Answer: B.
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