Particular Solutions: Initial Conditions - Calculus 2

1. The temperature $T(t)$ of a cooling object follows Newton's Law of Cooling: $\frac{dT}{dt} = -k(T-70)$ where the ambient temperature is 70°F. If $T(0) = 200°F$ and $T(10) = 150°F$, what is the temperature after 30 minutes?
   Answer: 98.8°F The general solution is $T(t) = 70 + Ce^{-kt}$. From $T(0) = 200$: $C = 130$. From $T(10) = 150$: $150 = 70 + 130e^{-10k}$, so $80 = 130e^{-10k}$, giving $e^{-10k} = \frac{8}{13}$. Therefore $e^{-k} = \left(\frac{8}{13}\right)^{1/10}$. At $t = 30$: $T(30) = 70 + 130\left(\frac{8}{13}\right)^3 = 70 + 130 \cdot \frac{512}{2197} \approx 98.8°F$. Choice A uses $k = 0.1$. Choice B results from incorrect exponentiation. Choice D comes from sign errors in the exponential.
2. The rate of change of a quantity $P$ is proportional to the square of $P$, i.e., $\frac{dP}{dt} = kP^2$. Given that $P(0) = 10$ and $P(1) = 5$, for what value of $t > 0$ is $P(t) = 2$?
   Answer: 4 Separating variables in $\frac{dP}{dt} = kP^2$ gives $P^{-2} dP = k dt$. Integrating yields $-P^{-1} = kt + C$, or $-\frac{1}{P} = kt + C$. Using $P(0)=10$, we get $-\frac{1}{10} = k(0) + C$, so $C = -\frac{1}{10}$. The equation becomes $-\frac{1}{P} = kt - \frac{1}{10}$. Using $P(1)=5$, we get $-\frac{1}{5} = k(1) - \frac{1}{10}$, which gives $-\frac{2}{10} = k - \frac{1}{10}$, so $k = -\frac{1}{10}$. The particular solution is $-\frac{1}{P} = -\frac{t}{10} - \frac{1}{10}$, which simplifies to $\frac{1}{P} = \frac{t+1}{10}$, or $P(t) = \frac{10}{t+1}$. We need to find $t$ when $P(t)=2$: $2 = \frac{10}{t+1} \implies 2(t+1) = 10 \implies t+1 = 5 \implies t=4$.
3. Given the differential equation $\frac{dy}{dx} = \frac{y}{x\ln(x)}$ and the initial condition $y(e)=e$, what is the value of $y(e^2)$?
   Answer: $2e$ Separating variables, we get $\frac{1}{y} dy = \frac{1}{x\ln(x)} dx$. Integrating both sides: $\int \frac{1}{y} dy = \ln|y|$. For the right side, we use u-substitution with $u = \ln(x)$ and $du = \frac{1}{x} dx$. So $\int \frac{1}{x\ln(x)} dx = \int \frac{1}{u} du = \ln|u| + C = \ln|\ln(x)| + C$. The general solution is $\ln|y| = \ln|\ln(x)| + C$. Using the initial condition $y(e)=e$: $\ln|e| = \ln|\ln(e)| + C$ $\implies 1 = \ln|1| + C \implies 1 = 0 + C \implies C=1$. The particular solution is $\ln|y| = \ln|\ln(x)| + 1$. Since $y(e)=e>0$, we can write $\ln(y) = \ln(\ln(x)) + 1$. To solve for $y$, we exponentiate: $y = e^{\ln(\ln x) + 1} = e^{\ln(\ln x)} \cdot e^1 = e \ln(x)$. Finally, we evaluate at $x=e^2$: $y(e^2) = e \ln(e^2) = e \cdot 2 = 2e$.
4. Find the particular solution to the differential equation $\frac{dy}{dx} = e^{x-y}$ that satisfies the initial condition $y(0) = \ln 3$. Then, evaluate this solution at $x=\ln 2$.
   Answer: $\ln 4$ The equation can be written as $\frac{dy}{dx} = e^x e^{-y}$. Separating the variables gives $e^y dy = e^x dx$. Integrating both sides yields $\int e^y dy = \int e^x dx$, which results in $e^y = e^x + C$. Using the initial condition $y(0) = \ln 3$, we have $e^{\ln 3} = e^0 + C$, which simplifies to $3 = 1 + C$, so $C=2$. The particular solution is $e^y = e^x + 2$. To find $y$ explicitly, we take the natural logarithm of both sides: $y = \ln(e^x + 2)$. Evaluating at $x=\ln 2$, we get $y = \ln(e^{\ln 2} + 2) = \ln(2+2) = \ln 4$.
5. Find the solution $y=f(x)$ to the differential equation $(x^2+1)y' = 2x(y-1)$ that passes through $(0, 2)$. What is $f(\sqrt{3})$?
   Answer: 5 First, we separate the variables: $\frac{dy}{y-1} = \frac{2x}{x^2+1} dx$. Integrating both sides gives $\int \frac{dy}{y-1} = \int \frac{2x}{x^2+1} dx$. The left side is $\ln|y-1|$. The right side can be solved with a u-substitution $u=x^2+1$, $du=2x dx$, so the integral is $\int \frac{du}{u} = \ln|u| = \ln(x^2+1)$ (since $x^2+1$ is always positive). So, $\ln|y-1| = \ln(x^2+1) + C$. Exponentiating, $|y-1| = e^C(x^2+1)$. Let $A=\pm e^C$, so $y-1 = A(x^2+1)$. Using the initial condition $y(0)=2$: $2-1 = A(0^2+1) \implies 1 = A$. The particular solution is $y-1 = 1(x^2+1)$, so $y = x^2+2$. Evaluating at $x=\sqrt{3}$, we get $f(\sqrt{3}) = (\sqrt{3})^2 + 2 = 3+2 = 5$.
6. Let $y(x)$ be the solution to the initial value problem $y' = y^2 - 4$ with $y(0)=3$. What is the value of $y(\frac{1}{4}\ln 2)$?
   Answer: $\frac{14}{3}$ We separate variables: $\frac{dy}{y^2-4} = dx$. We use partial fractions for the left side: $\frac{1}{(y-2)(y+2)} = \frac{A}{y-2} + \frac{B}{y+2}$. This gives $A=1/4$ and $B=-1/4$. The integral becomes $\frac{1}{4}\int (\frac{1}{y-2} - \frac{1}{y+2}) dy = \int dx$. This integrates to $\frac{1}{4}(\ln|y-2| - \ln|y+2|) = x+C$, or $\ln|\frac{y-2}{y+2}| = 4x+C_1$. Given $y(0)=3$, $\ln|\frac{3-2}{3+2}| = 0+C_1 \implies C_1=\ln(1/5)$. Since $y(0)=3>2$, the argument of the log is positive, so $\ln(\frac{y-2}{y+2}) = 4x+\ln(1/5)$. Exponentiating gives $\frac{y-2}{y+2} = e^{4x+\ln(1/5)} = \frac{1}{5}e^{4x}$. Let $x_0 = \frac{1}{4}\ln 2$. Then $e^{4x_0} = e^{\ln 2}=2$. So $\frac{y-2}{y+2} = \frac{2}{5}$. Solving for $y$: $5(y-2) = 2(y+2) \implies 5y-10 = 2y+4 \implies 3y=14 \implies y=14/3$.
7. The equation $\frac{dy}{dx} = \frac{y+1}{x}$ has a particular solution $y=f(x)$ with $f(1)=1$. What is $f(e)$?
   Answer: $2e-1$ Separating variables, we get $\frac{dy}{y+1} = \frac{dx}{x}$. Integrating both sides gives $\ln|y+1| = \ln|x| + C$. Exponentiating both sides gives $e^{\ln|y+1|} = e^{\ln|x|+C}$, which simplifies to $|y+1| = e^C |x|$. Let $A = \pm e^C$, so $y+1 = Ax$. Using the initial condition $y(1)=1$: $1+1 = A(1)$, so $A=2$. The particular solution is $y+1 = 2x$, or $y = 2x-1$. Evaluating at $x=e$, we find $f(e) = 2e-1$.
8. The rate of change of a population $P(t)$ is modeled by $\frac{dP}{dt} = kP(1000-P)$ where $k > 0$ is a constant. If $P(0) = 200$ and $P(2) = 400$, what is the population when $t = 4$?
   Answer: 667 individuals This is a logistic equation. Using separation of variables: $\frac{dP}{P(1000-P)} = k\,dt$. Partial fractions gives $\frac{1}{1000}\left(\frac{1}{P} + \frac{1}{1000-P}\right)dP = k\,dt$. Integrating and applying initial conditions $P(0) = 200$ and $P(2) = 400$ determines the constant. At $t = 4$, $P(4) = \frac{200000}{300} = 666.67 \approx 667$. Choice A assumes linear growth. Choice B results from incorrect partial fraction decomposition. Choice D comes from sign errors in the integration.
9. Let $y=f(x)$ be the particular solution to $\frac{dy}{dx} = (y-1)^2 \cos(x)$ with $f(0)=0$. What is the value of $f(\pi/2)$?
   Answer: 1/2 Separating variables gives $\frac{dy}{(y-1)^2} = \cos(x) dx$. Integrating both sides, $\int (y-1)^{-2} dy = \int \cos(x) dx$, we get $-(y-1)^{-1} = \sin(x) + C$, or $\frac{-1}{y-1} = \sin(x) + C$. Using the initial condition $y(0)=0$: $\frac{-1}{0-1} = \sin(0) + C \implies 1 = 0 + C \implies C=1$. The particular solution is $\frac{-1}{y-1} = \sin(x) + 1$. To find $f(\pi/2)$, we set $x=\pi/2$: $\frac{-1}{y-1} = \sin(\pi/2) + 1 = 1+1=2$. Solving for $y$: $-1 = 2(y-1) \implies -1 = 2y-2 \implies 1 = 2y \implies y=1/2$.
10. A function $y=f(x)$ is a solution to the differential equation $y' = xe^{-y}$. If the graph of the function passes through the point $(0, \ln 2)$, what is the value of $f(2)$?
    Answer: $\ln 4$ The differential equation is separable. We write $\frac{dy}{dx} = xe^{-y}$, which separates to $e^y dy = x dx$. Integrating both sides gives $\int e^y dy = \int x dx$, which results in $e^y = \frac{1}{2}x^2 + C$. Using the initial condition $y(0) = \ln 2$, we substitute $x=0$ and $y=\ln 2$: $e^{\ln 2} = \frac{1}{2}(0)^2 + C$, which simplifies to $2 = C$. The particular solution is $e^y = \frac{1}{2}x^2 + 2$. To find $f(2)$, we substitute $x=2$: $e^y = \frac{1}{2}(2)^2 + 2 = 2 + 2 = 4$. Solving for $y$ gives $y = \ln 4$.
11. The differential equation $\frac{dy}{dx} = \frac{xy}{x^2+1}$ has a particular solution passing through the point $(2, 3)$. What is the value of $y$ when $x = 4$?
    Answer: $3\sqrt{\frac{17}{5}}$ Separating variables: $\frac{dy}{y} = \frac{x\,dx}{x^2+1}$. Integrating: $\ln|y| = \frac{1}{2}\ln(x^2+1) + C$. This gives $y = K\sqrt{x^2+1}$. Using the initial condition $(2,3)$: $3 = K\sqrt{5}$, so $K = \frac{3}{\sqrt{5}}$. Therefore $y = \frac{3\sqrt{x^2+1}}{\sqrt{5}}$. When $x = 4$: $y = \frac{3\sqrt{17}}{\sqrt{5}} = 3\sqrt{\frac{17}{5}}$. Choice B squares the constant incorrectly. Choice C results from arithmetic errors. Choice D comes from incorrect integration of the right side.
12. The differential equation $xy' - 2y = xy$ describes a family of curves. Which curve in this family passes through the point $(1, e)$? Find the value of $y$ for this curve when $x=2$.
    Answer: $4e^2$ First, rearrange the differential equation to separate variables: $xy' = y(x+2) \implies y' = \frac{y(x+2)}{x}$. Now separate: $\frac{dy}{y} = \frac{x+2}{x} dx = (1 + \frac{2}{x}) dx$. Integrating both sides gives $\ln|y| = x + 2\ln|x| + C$. This can be rewritten as $\ln|y| = x + \ln(x^2) + C$. Using the initial condition $y(1)=e$: $\ln|e| = 1 + \ln(1^2) + C \implies 1 = 1 + 0 + C \implies C=0$. The particular solution is $\ln|y| = x + \ln(x^2)$. Since $y(1)=e>0$, we use $y$. To solve for $y$, we write $\ln(y) - \ln(x^2) = x \implies \ln(\frac{y}{x^2}) = x$. Exponentiating both sides gives $\frac{y}{x^2} = e^x$, so $y = x^2e^x$. At $x=2$, $y = 2^2 e^2 = 4e^2$.
13. A chemical reaction follows the rate law $\frac{dc}{dt} = -k c^2$ where $c(t)$ is the concentration at time $t$. If the initial concentration is $c(0) = 0.8$ M and the concentration after 10 minutes is $c(10) = 0.4$ M, what is the concentration after 25 minutes?
    Answer: 0.24 M This is a separable differential equation problem involving second-order kinetics. When you see a rate equation of the form $\frac{dc}{dt} = -kc^2$, you're dealing with a reaction where the rate depends on the square of the concentration. To solve this, separate variables: $\frac{dc}{c^2} = -k \, dt$. Integrating both sides gives $-\frac{1}{c} = -kt + C$. Using the initial condition $c(0) = 0.8$, we find $C = -\frac{1}{0.8} = -1.25$. So our solution is $\frac{1}{c(t)} = kt + 1.25$. Next, use the condition $c(10) = 0.4$ to find $k$. Substituting: $\frac{1}{0.4} = 10k + 1.25$, which gives $2.5 = 10k + 1.25$, so $k = 0.125$ min⁻¹. Now we can find $c(25)$: $\frac{1}{c(25)} = 0.125(25) + 1.25 = 4.375$. Therefore, $c(25) = \frac{1}{4.375} = 0.229$ M, which rounds to 0.24 M. Answer D (0.24 M) is correct. Answer A (0.20 M) likely comes from using first-order kinetics instead of second-order. Answer B (0.32 M) might result from incorrectly applying the half-life concept without accounting for the concentration dependence. Answer C (0.27 M) could arise from arithmetic errors in the integration or constant calculation. Remember: for second-order kinetics, the integrated rate law involves $\frac{1}{c}$, not $\ln c$ like first-order reactions. Always check which order kinetics you're dealing with first.
14. Let $y=f(x)$ be the particular solution to $\frac{dy}{dx} = \frac{2x+1}{2y}$ that passes through the point $(0, -2)$. Find $f(2)$.
    Answer: $-\sqrt{10}$ The equation separates to $2y \, dy = (2x+1) \, dx$. Integrating both sides gives $\int 2y \, dy = \int (2x+1) \, dx$, which results in $y^2 = x^2 + x + C$. Using the initial condition $y(0)=-2$, we find $C$: $(-2)^2 = 0^2 + 0 + C$, so $4=C$. The implicit particular solution is $y^2 = x^2 + x + 4$. Since the initial condition specifies $y=-2$ (a negative value), we must choose the negative branch of the solution: $y = -\sqrt{x^2+x+4}$. Evaluating at $x=2$, we find $f(2) = -\sqrt{2^2+2+4} = -\sqrt{4+2+4} = -\sqrt{10}$.
15. The rate of change of $y$ with respect to $x$ is inversely proportional to $y$ and directly proportional to $x$. The relationship is given by $\frac{dy}{dx} = \frac{kx}{y}$. If $y(0)=5$ and $y(4)=3$, what is the value of $k$?
    Answer: -1 Separating variables gives $y \, dy = kx \, dx$. Integrating both sides gives $\frac{1}{2}y^2 = \frac{k}{2}x^2 + C$, or $y^2 = kx^2 + C_1$ where $C_1=2C$. Using the first initial condition, $y(0)=5$: $5^2 = k(0)^2 + C_1 \implies 25 = C_1$. The equation becomes $y^2 = kx^2 + 25$. Using the second condition, $y(4)=3$: $3^2 = k(4)^2 + 25 \implies 9 = 16k + 25$. Solving for $k$: $16k = 9 - 25 = -16$, so $k = -1$.
16. A radioactive substance decays according to $\frac{dy}{dt} = -ky$ where $y(t)$ is the amount remaining at time $t$ hours. If the half-life is 8 hours and $y(4) = 30$ grams, what was the initial amount $y(0)$?
    Answer: 42.43 grams From the half-life, $k = \frac{\ln 2}{8}$. The general solution is $y(t) = Ce^{-kt}$. Given $y(4) = 30$, we have $30 = Ce^{-4k} = Ce^{-4\ln 2/8} = Ce^{-\ln 2/2} = C\cdot 2^{-1/2} = \frac{C}{\sqrt{2}}$. Therefore $C = 30\sqrt{2} = 42.43$. Choice A uses $k = 8$ instead of $k = \ln 2/8$. Choice C incorrectly applies the half-life formula. Choice D results from using $e^{\ln 2/2}$ instead of $e^{-\ln 2/2}$.
17. Consider the differential equation $\frac{dy}{dx} = \frac{\cos(x)}{y^2}$. If $y = 3$ when $x = \frac{\pi}{2}$, what is the value of $y^3$ when $x=0$?
    Answer: 24 Separating variables gives $y^2 dy = \cos(x) dx$. Integrating both sides, we get $\int y^2 dy = \int \cos(x) dx$, which yields $\frac{1}{3}y^3 = \sin(x) + C$. Using the initial condition $y(\pi/2) = 3$, we substitute to find C: $\frac{1}{3}(3)^3 = \sin(\frac{\pi}{2}) + C$, which simplifies to $9 = 1 + C$, so $C=8$. The particular solution is $\frac{1}{3}y^3 = \sin(x) + 8$, or $y^3 = 3\sin(x) + 24$. When $x=0$, $y^3 = 3\sin(0) + 24 = 3(0) + 24 = 24$.
18. Let $y=f(x)$ be the particular solution to the differential equation $\frac{dy}{dx} = \frac{3x^2}{2y}$ with the initial condition $f(2)=3$. What is the value of $f(3)$?
    Answer: $2\sqrt{7}$ Separating variables gives $2y \, dy = 3x^2 \, dx$. Integrating both sides yields $\int 2y \, dy = \int 3x^2 \, dx$, which results in $y^2 = x^3 + C$. Using the initial condition $y(2)=3$, we have $3^2 = 2^3 + C$, so $9 = 8 + C$, which means $C=1$. The particular solution is $y^2 = x^3 + 1$. Since the initial value of $y$ is positive (3), we take the positive square root: $y = \sqrt{x^3 + 1}$. Evaluating at $x=3$, we get $y = \sqrt{3^3 + 1} = \sqrt{27 + 1} = \sqrt{28} = 2\sqrt{7}$.
19. Consider the initial value problem $y' = 1 + y^2$ with $y(0)=1$. What is the value of $y(\pi/6)$?
    Answer: $2+\sqrt{3}$ Separating variables gives $\frac{dy}{1+y^2} = dx$. Integrating both sides yields $\arctan(y) = x + C$. Using the initial condition $y(0)=1$, we get $\arctan(1) = 0 + C$, which means $C = \pi/4$. The particular solution is $\arctan(y) = x + \pi/4$, so $y = \tan(x+\pi/4)$. We need to evaluate $y(\pi/6)$: $y = \tan(\frac{\pi}{6} + \frac{\pi}{4}) = \tan(\frac{2\pi+3\pi}{12}) = \tan(\frac{5\pi}{12})$. Using the tangent addition formula: $\tan(75°) = \tan(45° + 30°) = \frac{\tan 45° + \tan 30°}{1 - \tan 45° \tan 30°} = \frac{1 + \frac{1}{\sqrt{3}}}{1 - \frac{1}{\sqrt{3}}} = \frac{\sqrt{3}+1}{\sqrt{3}-1} = \frac{(\sqrt{3}+1)^2}{2} = 2+\sqrt{3}$.

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