Solving Separable DEs - Calculus 2

The solution to $y' = y \cos(x)$ satisfies $y(0) = 1$ . Evaluate $y(\pi/2)$ .
Answer: $e$ When you encounter a differential equation like $y' = y \cos(x)$ , you're dealing with a separable differential equation. The key insight is recognizing that you can separate the variables and integrate both sides. Start by rewriting the equation as $\frac{dy}{dx} = y \cos(x)$ , then separate variables: $\frac{dy}{y} = \cos(x) dx$ . Now integrate both sides: $\int \frac{dy}{y} = \int \cos(x) dx$ , which gives you $\ln|y| = \sin(x) + C$ . Exponentiating both sides yields $y = Ae^{\sin(x)}$ where $A = e^C$ . Use the initial condition $y(0) = 1$ to find $A$ : $1 = Ae^{\sin(0)} = Ae^0 = A$ , so $A = 1$ . Therefore, $y = e^{\sin(x)}$ . To find $y(\pi/2)$ , substitute: $y(\pi/2) = e^{\sin(\pi/2)} = e^1 = e$ . Looking at the wrong answers: Choice A, $e^{\pi/2}$ , results from incorrectly integrating $\cos(x)$ as $x$ instead of $\sin(x)$ . Choice B, $1$ , comes from forgetting to apply the exponential function after integration or assuming the solution is constant. Choice D, $0$ , might stem from confusing this with a different type of differential equation or making algebraic errors. The correct answer is C, $e$ . Remember: for separable differential equations, always separate variables completely before integrating, and don't forget to use your initial conditions to determine any constants. Practice recognizing when $\sin(\pi/2) = 1$ to avoid computational errors.
A particle moves along the y-axis such that its velocity is given by $v(t) = \frac{dy}{dt} = 2t(y+1)$ . If $y(1)=e-1$ , find $y(2)$ .
Answer: $e^4 - 1$ This is a separable differential equation problem where you need to solve for the particle's position given its velocity function and an initial condition. When you see $\frac{dy}{dt} = 2t(y+1)$ , recognize this as separable since you can isolate all y terms on one side and all t terms on the other. Separate the variables: $\frac{dy}{y+1} = 2t \, dt$ Integrate both sides: $\int \frac{dy}{y+1} = \int 2t \, dt$ This gives you: $\ln|y+1| = t^2 + C$ Exponentiate both sides: $|y+1| = e^{t^2 + C} = Ae^{t^2}$ where $A = e^C$ Since we're dealing with position, assume $y+1 > 0$ , so: $y+1 = Ae^{t^2}$ , which means $y = Ae^{t^2} - 1$ Use the initial condition $y(1) = e-1$ to find A: $e-1 = Ae^{1^2} - 1 = Ae - 1$ Solving: $e = Ae$ , so $A = 1$ Therefore: $y(t) = e^{t^2} - 1$ At $t = 2$ : $y(2) = e^{2^2} - 1 = e^4 - 1$ Choice A ( $e^3 - 1$ ) likely comes from incorrectly computing $2^2 = 2 \times 2$ as just $3$ . Choice C ( $e^4$ ) results from forgetting the $-1$ constant term. Choice D ( $2e - 1$ ) suggests confusion with the coefficient in the original equation. Strategy tip: In separable differential equations, always check your constant of integration using the initial condition, and be careful with exponent arithmetic when substituting your final t-value.
Consider the differential equation $\frac{dy}{dx} = \frac{x^2 + 1}{y^2 - 4}$ with initial condition $y(0) = 3$ . Which of the following represents the solution in implicit form?
Answer: $\frac{y^3}{3} - 4y = \frac{x^3}{3} + x - 3$ When you encounter a differential equation where variables can be separated, your goal is to move all terms involving one variable to one side and all terms involving the other variable to the opposite side, then integrate both sides. Starting with $\frac{dy}{dx} = \frac{x^2 + 1}{y^2 - 4}$ , you can separate variables by multiplying both sides by $(y^2 - 4)$ and $dx$ : $(y^2 - 4)dy = (x^2 + 1)dx$ Now integrate both sides. The left side gives: $\int (y^2 - 4)dy = \frac{y^3}{3} - 4y + C_1$ The right side gives: $\int (x^2 + 1)dx = \frac{x^3}{3} + x + C_2$ Setting them equal: $\frac{y^3}{3} - 4y = \frac{x^3}{3} + x + C$ where $C = C_2 - C_1$ To find $C$ , use the initial condition $y(0) = 3$ : $\frac{3^3}{3} - 4(3) = \frac{0^3}{3} + 0 + C$ , which gives $9 - 12 = C$ , so $C = -3$ Therefore, the solution is $\frac{y^3}{3} - 4y = \frac{x^3}{3} + x - 3$ , making D correct. A and C both have $C = +3$ instead of $C = -3$ , showing incorrect application of the initial condition. B has $C = -\frac{1}{3}$ , which would result from arithmetic errors when substituting the initial condition. Remember: after separating variables and integrating, always substitute your initial condition carefully to find the constant. Double-check your arithmetic since sign errors are common in these problems.
A population of bacteria grows according to the differential equation $\frac{dP}{dt} = kP(1000 - P)$ , where $P(t)$ is the population at time $t$ and $k > 0$ is a constant. If the initial population is $P(0) = 50$ and the population reaches 200 after 2 hours, what is the population after 4 hours?
Answer: $\frac{16000}{41}$ bacteria This is a separable differential equation. Separating variables: $\frac{dP}{P(1000-P)} = k\,dt$ . Using partial fractions: $\frac{1}{P(1000-P)} = \frac{1}{1000}\left(\frac{1}{P} + \frac{1}{1000-P}\right)$ . Integrating: $\frac{1}{1000}\ln\left|\frac{P}{1000-P}\right| = kt + C$ . With $P(0) = 50$ : $C = \frac{1}{1000}\ln\left(\frac{50}{950}\right) = \frac{1}{1000}\ln\left(\frac{1}{19}\right)$ . Using $P(2) = 200$ : $\frac{1}{1000}\ln\left(\frac{200}{800}\right) = 2k + \frac{1}{1000}\ln\left(\frac{1}{19}\right)$ , which gives $k = \frac{1}{2000}\ln\left(\frac{19}{4}\right)$ . For $t = 4$ : $\ln\left(\frac{P}{1000-P}\right) = \ln\left(\frac{19}{16}\right)$ , so $\frac{P}{1000-P} = \frac{19}{16}$ . Solving: $P = \frac{16000}{41}$ .
Let $y=f(x)$ be the solution to the initial value problem $y' + 2xy = 0$ with $y(0) = e$ . Find the positive value of $x$ for which $y(x) = 1$ .
Answer: $1$ First, rewrite the equation as $\frac{dy}{dx} = -2xy$ . Separate the variables: $\frac{1}{y} dy = -2x dx$ . Integrate both sides: $\ln|y| = -x^2 + C$ . Use the initial condition $y(0) = e$ : $\ln|e| = -0^2 + C \implies 1 = C$ . The particular solution is $\ln|y| = -x^2 + 1$ . Since $y(0)=e>0$ , we can write $\ln(y) = 1 - x^2$ . The explicit solution is $y = e^{1-x^2}$ . Now, set $y(x) = 1$ : $1 = e^{1-x^2}$ . Taking the natural logarithm of both sides gives $0 = 1 - x^2$ , so $x^2 = 1$ . The positive value is $x=1$ .
Let $y=f(x)$ be the solution for $y' = x e^{-y}$ with $f(1)=0$ . What is the value of $f(3)$ ?
Answer: $\ln(5)$ This is a separable differential equation problem where you need to isolate variables and integrate both sides to find the particular solution. When you see $y' = xe^{-y}$ , recognize this as separable since you can write it as $\frac{dy}{dx} = xe^{-y}$ . Multiply both sides by $e^y$ and $dx$ to get $e^y \, dy = x \, dx$ . Now integrate both sides: $\int e^y \, dy = \int x \, dx$ , which gives you $e^y = \frac{x^2}{2} + C$ . Use the initial condition $f(1) = 0$ to find $C$ . When $x = 1$ and $y = 0$ : $e^0 = \frac{1^2}{2} + C$ , so $1 = \frac{1}{2} + C$ , meaning $C = \frac{1}{2}$ . Your general solution is $e^y = \frac{x^2}{2} + \frac{1}{2} = \frac{x^2 + 1}{2}$ . To find $f(3)$ , substitute $x = 3$ : $e^y = \frac{9 + 1}{2} = 5$ . Taking the natural logarithm: $y = \ln(5)$ , which is answer choice D. Answer A, $\ln(11/2)$ , would result from incorrectly adding instead of finding $C = \frac{1}{2}$ . Answer B, $\ln(9/2)$ , comes from forgetting to add the constant $C$ entirely. Answer C, $5$ , is the trap of stopping at $e^y = 5$ without taking the logarithm. Remember: with separable equations, always separate variables completely before integrating, and don't forget to apply initial conditions to find your constant of integration.
A curve $y=f(x)$ passes through the point $(e, 2)$ and its slope at any point $(x,y)$ is given by $\frac{dy}{dx} = \frac{y}{x \ln(x)}$ . What is the value of $y$ when $x=e^4$ ?
Answer: $8$ Separate the variables: $\frac{1}{y} dy = \frac{1}{x \ln(x)} dx$ . Integrate both sides. The left side is $\ln|y|$ . For the right side, use u-substitution with $u=\ln(x)$ and $du=\frac{1}{x}dx$ . This gives $\int \frac{1}{u} du = \ln|u| = \ln|\ln(x)|$ . So, $\ln|y| = \ln|\ln(x)| + C$ . Using the point $(e,2)$ : $\ln(2) = \ln|\ln(e)| + C \implies \ln(2) = \ln(1) + C \implies C = \ln(2)$ . The solution is $\ln|y| = \ln|\ln(x)| + \ln(2) = \ln(2|\ln(x)|)$ . So, $|y| = 2|\ln(x)|$ . Since $y(e)=2>0$ and for $x>1$ , $\ln(x)>0$ , we have $y = 2\ln(x)$ . At $x=e^4$ , $y = 2\ln(e^4) = 2 \cdot 4 = 8$ .
Let $y(t)$ be the solution to the initial value problem $\frac{dy}{dt} = 2y(1 - \frac{y}{8})$ with $y(0) = 2$ . Find $\lim_{t\to\infty} y(t)$ .
Answer: $8$ The differential equation is a logistic growth model of the form $\frac{dy}{dt} = ky(1 - \frac{y}{K})$ , where $K$ is the carrying capacity. By comparing the given equation with the standard form, we can identify the carrying capacity $K=8$ . For any initial condition $y(0) > 0$ , the solution $y(t)$ will approach the carrying capacity $K$ as $t \to \infty$ . Since $y(0) = 2 > 0$ , $\lim_{t\to\infty} y(t) = 8$ .
The rate of change of a quantity $P$ is directly proportional to the square root of $P$ . If $P=9$ when $t=1$ and $P=25$ when $t=2$ , what is $P$ when $t=0$ ?
Answer: $1$ When you encounter a problem stating that "the rate of change of a quantity is directly proportional to" something, you're dealing with a differential equation. The phrase "rate of change" means derivative, so you need to set up and solve $\frac{dP}{dt} = k\sqrt{P}$ where $k$ is the constant of proportionality. To solve this separable differential equation, separate variables: $\frac{dP}{\sqrt{P}} = k \, dt$ . Integrating both sides gives $2\sqrt{P} = kt + C$ , or $\sqrt{P} = \frac{kt + C}{2}$ . Therefore, $P = \left(\frac{kt + C}{2}\right)^2$ . Using the given conditions, when $t = 1$ , $P = 9$ : $9 = \left(\frac{k + C}{2}\right)^2$ , so $\frac{k + C}{2} = 3$ , giving us $k + C = 6$ . When $t = 2$ , $P = 25$ : $25 = \left(\frac{2k + C}{2}\right)^2$ , so $\frac{2k + C}{2} = 5$ , giving us $2k + C = 10$ . Solving this system: From the second equation minus the first, $k = 4$ . Substituting back: $C = 2$ . Therefore, $P = \left(\frac{4t + 2}{2}\right)^2 = (2t + 1)^2$ . When $t = 0$ : $P = (2(0) + 1)^2 = 1$ , confirming answer C. Answer A ( $P = 0$ ) would require $C = 0$ , which contradicts our conditions. Answer B ( $P = 4$ ) and D ( $P = 2$ ) result from arithmetic errors in solving the system of equations or incorrectly setting up the integration constant. Study tip: Always use both given conditions to find both unknown constants, and verify your final equation works for the given data points.
Solve the initial value problem $\frac{dy}{d\theta} = \sec^2(\theta) \sec(y)$ with the initial condition $y(\pi/4) = \pi/6$ . Find the value of $y(0)$ .
Answer: $-\pi/6$ Rewrite the equation as $\frac{dy}{d\theta} = \frac{\sec^2(\theta)}{\cos(y)}$ . Separate the variables: $\cos(y) dy = \sec^2(\theta) d\theta$ . Integrate both sides: $\int \cos(y) dy = \int \sec^2(\theta) d\theta$ , which gives $\sin(y) = \tan(\theta) + C$ . Use the initial condition $y(\pi/4) = \pi/6$ : $\sin(\pi/6) = \tan(\pi/4) + C \implies 1/2 = 1 + C \implies C = -1/2$ . The particular solution is $\sin(y) = \tan(\theta) - 1/2$ . To find $y(0)$ , substitute $\theta=0$ : $\sin(y) = \tan(0) - 1/2 = -1/2$ . Therefore, $y = \arcsin(-1/2) = -\pi/6$ .
The solution to the differential equation $\frac{dy}{dx} = \frac{xy + x}{y^2 + 1}$ passes through the point $(2, 1)$ . Which of the following equations implicitly defines this solution?
Answer: $y^2 - 2y - x^2 + 4\ln(y+1) = 4\ln(2) - 5$ First, factor the numerator: $\frac{dy}{dx} = \frac{x(y+1)}{y^2+1}$ . Separate variables: $\frac{y^2+1}{y+1} dy = x dx$ . Use polynomial long division or substitution on the left side: $\frac{y^2-1+2}{y+1} = \frac{(y-1)(y+1)+2}{y+1} = y-1 + \frac{2}{y+1}$ . Now integrate: $\int (y-1 + \frac{2}{y+1}) dy = \int x dx$ . This gives $\frac{y^2}{2} - y + 2\ln|y+1| = \frac{x^2}{2} + C$ . Use the point $(2,1)$ : $\frac{1^2}{2} - 1 + 2\ln|1+1| = \frac{2^2}{2} + C \implies -\frac{1}{2} + 2\ln(2) = 2 + C \implies C = 2\ln(2) - \frac{5}{2}$ . The implicit solution is $\frac{y^2}{2} - y + 2\ln|y+1| = \frac{x^2}{2} + 2\ln(2) - \frac{5}{2}$ . Since $y(2)=1$ , $y+1>0$ near the initial condition. Multiply by 2 to clear fractions: $y^2 - 2y + 4\ln(y+1) = x^2 + 4\ln(2) - 5$ , which rearranges to $y^2 - 2y - x^2 + 4\ln(y+1) = 4\ln(2) - 5$ .
Find the particular solution to $x\frac{dy}{dx} = 2(y-5)$ that satisfies the initial condition $y(1) = 10$ .
Answer: $y = 5x^2 + 5$ When you encounter a differential equation like $x\frac{dy}{dx} = 2(y-5)$ , you're dealing with a separable differential equation. The key is to separate variables so that all terms involving $y$ are on one side and all terms involving $x$ are on the other. Start by rearranging: $\frac{dy}{dx} = \frac{2(y-5)}{x}$ . Now separate variables: $\frac{dy}{y-5} = \frac{2dx}{x}$ . Integrate both sides: $\int \frac{dy}{y-5} = \int \frac{2dx}{x}$ . This gives you $\ln|y-5| = 2\ln|x| + C = \ln(x^2) + C$ . Exponentiating both sides: $|y-5| = e^{\ln(x^2) + C} = Ax^2$ where $A = e^C$ . Since we can absorb the absolute value into the constant, we get $y-5 = Ax^2$ , so $y = Ax^2 + 5$ . Apply the initial condition $y(1) = 10$ : $10 = A(1)^2 + 5$ , which gives $A = 5$ . Therefore, $y = 5x^2 + 5$ , which is choice B. Choice A incorrectly uses $e^{x^2}$ instead of $x^2$ - this would come from mishandling the integration. Choice C has the wrong constant term (9 instead of 5) and coefficient, suggesting errors in applying the initial condition. Choice D uses $\ln(x^2)$ directly without exponentiating, which represents an incomplete solution to the differential equation. Remember: for separable equations, always separate variables completely before integrating, and don't forget to apply your initial conditions to find the specific constant.
The solution $y(x)$ to $\frac{dy}{dx} = \frac{2y}{x}$ satisfies $y(-1) = -3$ . Find the value of $y(-e)$ .
Answer: $-3e^2$ When you encounter a differential equation of the form $\frac{dy}{dx} = \frac{ky}{x}$ , you're dealing with a separable equation that requires careful handling of the initial condition and domain considerations. First, separate variables: $\frac{dy}{y} = \frac{2dx}{x}$ . Integrating both sides gives $\ln|y| = 2\ln|x| + C$ , which simplifies to $|y| = A|x|^2$ where $A = e^C$ . Since we need $y(x) = \pm A|x|^2$ , we must determine the sign using our initial condition. With $y(-1) = -3$ , we get $-3 = \pm A|-1|^2 = \pm A(1)$ , so $A = 3$ and we need the negative branch: $y(x) = -3|x|^2$ . For $x < 0$ (our domain of interest), $|x| = -x$ , so $y(x) = -3(-x)^2 = -3x^2$ . Therefore, $y(-e) = -3(-e)^2 = -3e^2$ . Let's examine why the other answers are incorrect: **A) $3/e^2$ **: This has the wrong sign (positive instead of negative) and inverts the exponential term. **B) $3e^2$ **: This only has the wrong sign—it's the positive version of our answer. **C) $-3$ **: This ignores the $x^2$ factor entirely, treating the solution as constant. Strategy tip: When solving separable differential equations with initial conditions in different domains (positive vs. negative $x$ ), always check how absolute value expressions behave in your specific region. The sign of your solution often depends on which branch you're working with.
The solution to $(1+x^2)y' - 2xy^2 = 0$ passes through the point $(0, 1)$ . What is the value of $y(1)$ ?
Answer: $\frac{1}{1 - \ln(2)}$ When you encounter a differential equation like $(1+x^2)y' - 2xy^2 = 0$ , recognize this as a separable differential equation. The key is to rearrange terms so all $y$ terms are on one side and all $x$ terms are on the other. First, rewrite the equation: $(1+x^2)y' = 2xy^2$ , so $y' = \frac{2xy^2}{1+x^2}$ Since $y' = \frac{dy}{dx}$ , we can separate variables: $\frac{dy}{y^2} = \frac{2x}{1+x^2}dx$ Integrating both sides: $\int y^{-2}dy = \int \frac{2x}{1+x^2}dx$ The left side gives $-\frac{1}{y}$ . For the right side, use substitution $u = 1+x^2$ , so $du = 2x dx$ : $\int \frac{2x}{1+x^2}dx = \int \frac{du}{u} = \ln|u| = \ln(1+x^2)$ Therefore: $-\frac{1}{y} = \ln(1+x^2) + C$ Using the initial condition $(0,1)$ : $-\frac{1}{1} = \ln(1) + C$ , so $C = -1$ . The solution becomes: $-\frac{1}{y} = \ln(1+x^2) - 1$ Solving for $y$ : $y = \frac{-1}{\ln(1+x^2) - 1} = \frac{1}{1 - \ln(1+x^2)}$ At $x = 1$ : $y(1) = \frac{1}{1 - \ln(2)}$ This confirms answer D. Answers A, B, and C likely result from integration errors, incorrect separation of variables, or sign mistakes during algebraic manipulation. Strategy tip: Always verify your solution satisfies both the differential equation and initial condition—this catches most algebraic errors in separable differential equations.
What is the particular solution to the differential equation $xy' = \sqrt{y}$ given the initial condition $y(1) = 4$ ?
Answer: $y = \frac{1}{4}(\ln(x) + 4)^2$ This is a separable differential equation, which means you can isolate all terms with $y$ on one side and all terms with $x$ on the other. When you see $xy' = \sqrt{y}$ , rewrite $y'$ as $\frac{dy}{dx}$ to get $x\frac{dy}{dx} = \sqrt{y}$ . To separate variables, divide both sides by $\sqrt{y}$ and multiply by $dx$ : $\frac{dy}{\sqrt{y}} = \frac{dx}{x}$ . Now integrate both sides. The left side becomes $\int y^{-1/2} dy = 2\sqrt{y}$ , and the right side becomes $\int \frac{dx}{x} = \ln(x) + C$ . This gives us $2\sqrt{y} = \ln(x) + C$ . Using the initial condition $y(1) = 4$ : $2\sqrt{4} = \ln(1) + C$ , so $4 = 0 + C$ , meaning $C = 4$ . Therefore, $2\sqrt{y} = \ln(x) + 4$ . Solving for $y$ : $\sqrt{y} = \frac{\ln(x) + 4}{2}$ , so $y = \left(\frac{\ln(x) + 4}{2}\right)^2 = \frac{1}{4}(\ln(x) + 4)^2$ . Answer choice A incorrectly assumes $C = 2$ instead of $C = 4$ . Choice B treats this as a first-order linear equation rather than separable, leading to an incorrect form. Choice C has the wrong constant and missing the factor of $\frac{1}{4}$ that comes from squaring the fraction. When solving separable differential equations, always check your constant of integration carefully using the initial condition, and be meticulous about algebraic manipulation when solving for the dependent variable.
A particle moves along a line such that its velocity $v(t)$ satisfies $\frac{dv}{dt} = -2tv^2$ with $v(0) = \frac{1}{3}$ . At what time does the velocity reach $\frac{1}{6}$ ?
Answer: $t = \sqrt{3}$ Separating variables: $\frac{dv}{v^2} = -2t \, dt$ . Integrating: $\int v^{-2} dv = \int -2t \, dt$ gives $-\frac{1}{v} = -t^2 + C$ . Using $v(0) = \frac{1}{3}$ : $-3 = 0 + C$ , so $C = -3$ . Therefore $-\frac{1}{v} = -t^2 - 3$ , which gives $\frac{1}{v} = t^2 + 3$ , so $v = \frac{1}{t^2 + 3}$ . When $v = \frac{1}{6}$ : $\frac{1}{6} = \frac{1}{t^2 + 3}$ , so $t^2 + 3 = 6$ , giving $t^2 = 3$ and $t = \sqrt{3}$ (taking the positive root since we're looking for a future time).
A solution to $\frac{dy}{dx} = \frac{3x^2+1}{2y}$ passes through the point $(1, 2)$ . What is the value of $y$ when $x=0$ ?
Answer: $\sqrt{2}$ This is a separable differential equation problem where you need to find a particular solution using an initial condition. When you see $\frac{dy}{dx} = \frac{f(x)}{g(y)}$ , you can separate variables by multiplying both sides by $g(y)dx$ , giving you $g(y)dy = f(x)dx$ . Starting with $\frac{dy}{dx} = \frac{3x^2+1}{2y}$ , multiply both sides by $2y dx$ to get $2y dy = (3x^2+1)dx$ . Now integrate both sides: $\int 2y dy = \int (3x^2+1)dx$ , which gives you $y^2 = x^3 + x + C$ . To find the constant $C$ , use the initial condition that the solution passes through $(1, 2)$ . Substituting: $2^2 = 1^3 + 1 + C$ , so $4 = 2 + C$ , meaning $C = 2$ . Your particular solution is $y^2 = x^3 + x + 2$ . When $x = 0$ : $y^2 = 0^3 + 0 + 2 = 2$ , so $y = \sqrt{2}$ , which is answer C. Looking at the wrong answers: A) $\sqrt{3}$ would result from incorrectly calculating $C = 3$ instead of $C = 2$ . B) $2$ is the y-value from the initial condition, which some students mistakenly think remains constant. D) $0$ would come from forgetting to include the constant of integration entirely. Remember: with separable differential equations, always separate variables first, integrate both sides, then use the initial condition to find your constant. The initial condition gives you one point, but you need it to find the entire solution curve.
Find the particular solution $y=f(x)$ to the differential equation $y' = y^2 / x$ with the initial condition $y(1) = -1/2$ . On what largest open interval is this solution defined?
Answer: $(e^{-2}, \infty)$ Separate the variables: $\frac{1}{y^2} dy = \frac{1}{x} dx$ . Integrate both sides: $-\frac{1}{y} = \ln|x| + C$ . Use the initial condition $y(1) = -1/2$ : $-\frac{1}{-1/2} = \ln|1| + C \implies 2 = 0 + C \implies C = 2$ . The particular solution is $-\frac{1}{y} = \ln|x| + 2$ . Since the initial condition is at $x=1>0$ , we can write $-\frac{1}{y} = \ln(x) + 2$ . Solving for $y$ gives $y = \frac{-1}{\ln(x) + 2}$ . The solution is defined when $x>0$ (due to $\ln(x)$ ) and when the denominator is not zero. $\ln(x) + 2 \neq 0 \implies \ln(x) \neq -2 \implies x \neq e^{-2}$ . The domain is $(0, e^{-2}) \cup (e^{-2}, \infty)$ . The initial condition $x=1$ is in the interval $(e^{-2}, \infty)$ because $e^{-2} \approx 0.135 < 1$ . Therefore, the largest open interval containing $x=1$ where the solution is defined is $(e^{-2}, \infty)$ .
Consider the differential equation $\frac{dy}{dx} = \frac{\cos(x)}{y^2}$ with the initial condition $y(\pi/2) = 3$ . What is the value of $y(0)$ ?
Answer: $2 \sqrt[3]{3}$ First, separate the variables: $y^2 dy = \cos(x) dx$ . Integrating both sides gives $\frac{y^3}{3} = \sin(x) + C$ . Use the initial condition $y(\pi/2) = 3$ to find $C$ : $\frac{3^3}{3} = \sin(\pi/2) + C \implies 9 = 1 + C \implies C = 8$ . The particular solution is $\frac{y^3}{3} = \sin(x) + 8$ , or $y^3 = 3\sin(x) + 24$ . To find $y(0)$ , substitute $x=0$ : $y^3 = 3\sin(0) + 24 = 24$ . Thus, $y(0) = \sqrt[3]{24} = \sqrt[3]{8 \cdot 3} = 2\sqrt[3]{3}$ .
Find the implicit solution to the differential equation $(y-1)\frac{dy}{dx} = x^2$ with the initial condition $y(3)=1$ .
Answer: $y^2 - 2y - 2x^3/3 = -19$ When you encounter a separable differential equation like this, your goal is to separate variables and integrate both sides, then apply the initial condition to find the constant of integration. Starting with $(y-1)\frac{dy}{dx} = x^2$ , you can separate variables by moving all terms with $y$ to one side and all terms with $x$ to the other: $(y-1)dy = x^2 dx$ . Now integrate both sides. The left side gives you $\int (y-1)dy = \frac{y^2}{2} - y$ , and the right side gives you $\int x^2 dx = \frac{x^3}{3}$ . This yields $\frac{y^2}{2} - y = \frac{x^3}{3} + C$ . Multiplying through by 2 to clear fractions: $y^2 - 2y = \frac{2x^3}{3} + 2C$ . Let $K = 2C$ , so $y^2 - 2y - \frac{2x^3}{3} = K$ . Apply the initial condition $y(3) = 1$ : $1^2 - 2(1) - \frac{2(3)^3}{3} = K$ , which gives $1 - 2 - \frac{54}{3} = 1 - 2 - 18 = -19$ . Therefore, the solution is $y^2 - 2y - \frac{2x^3}{3} = -19$ , which is answer B. Answer A has the wrong constant (-1 instead of -19). Answer C uses $x^3$ instead of $\frac{2x^3}{3}$ and has the wrong constant. Answer D appears to use substitution methods inappropriately—this equation doesn't require logarithmic techniques since it's directly separable. Always apply initial conditions after finding the general solution to determine the specific constant value. Double-check your integration, especially with fractions.

Start this quiz