General Solutions: Separation of Variables - Calculus 2

1. The general solution to the differential equation $\frac{dy}{dx} = \frac{1+y^2}{\sqrt{1-x^2}}$ is which of the following?
   Answer: y = \tan(\arcsin(x) + C) Separate the variables to get $\frac{1}{1+y^2} dy = \frac{1}{\sqrt{1-x^2}} dx$. Integrate both sides: $\int \frac{1}{1+y^2} dy = \int \frac{1}{\sqrt{1-x^2}} dx$. Both are standard inverse trigonometric integrals. The result is $\arctan(y) = \arcsin(x) + C$. To solve for $y$, take the tangent of both sides: $y = \tan(\arcsin(x) + C)$.
2. What is the general solution for the differential equation $y' \sin(y) = \cos(x) - \sin(x)$?
   Answer: -\cos(y) = \sin(x) + \cos(x) + C This is a separable differential equation, which means you can separate the variables and integrate both sides. When you see an equation where you can get all terms involving $y$ on one side and all terms involving $x$ on the other, separation of variables is your go-to method. Starting with $y' \sin(y) = \cos(x) - \sin(x)$, rewrite $y'$ as $\frac{dy}{dx}$ and separate: $\sin(y) \, dy = (\cos(x) - \sin(x)) \, dx$. Now integrate both sides. The left side gives $\int \sin(y) \, dy = -\cos(y)$. The right side gives $\int (\cos(x) - \sin(x)) \, dx = \sin(x) - (-\cos(x)) = \sin(x) + \cos(x)$. Therefore: $-\cos(y) = \sin(x) + \cos(x) + C$, which is answer D. Answer A incorrectly uses $\frac{1}{2}\sin^2(y)$ as the antiderivative of $\sin(y)$, but the correct antiderivative is $-\cos(y)$. Answer B has the correct form for the right side but uses $\cos(y)$ instead of $-\cos(y)$, missing the negative sign from integration. Answer C gets the left side correct as $-\cos(y)$, but incorrectly integrates the right side as $-\sin(x) - \cos(x)$, putting negative signs where they don't belong. Remember: when integrating $\sin(y)$, you get $-\cos(y)$, and when integrating $\cos(x) - \sin(x)$, you get $\sin(x) + \cos(x)$. Double-check your antiderivatives—they're where most errors occur in separable equations.
3. A population model is described by the differential equation $\frac{dP}{dt} = \frac{P(100-P)}{t+1}$, where $P(t)$ represents population at time $t$. After separating variables, what is the correct integral setup?
   Answer: $\int \frac{dP}{P(100-P)} = \int \frac{dt}{t+1}$ Separating variables: $\frac{dP}{P(100-P)} = \frac{dt}{t+1}$. Integrating both sides gives the setup in choice A. Choice B fails to separate properly. Choice C incorrectly splits the left integral (partial fractions comes after separation). Choice D incorrectly moves only part of the denominator.
4. Which of the following implicitly defines the general solution to $\frac{dy}{dx} = \frac{\ln(x)}{xy}$?
   Answer: $y^2$ = (\ln x)^2 + C To solve, we separate the variables: $y dy = \frac{\ln(x)}{x} dx$. Next, we integrate both sides: $\int y dy = \int \frac{\ln(x)}{x} dx$. The left side is $\frac{y^2}{2}$. For the right side, we use u-substitution with $u = \ln(x)$, so $du = \frac{1}{x} dx$. The integral becomes $\int u du = \frac{u^2}{2} = \frac{(\ln x)^2}{2}$. So, we have $\frac{y^2}{2} = \frac{(\ln x)^2}{2} + C_1$. Multiplying by 2 gives $y^2 = (\ln x)^2 + 2C_1$. Letting $C = 2C_1$ gives the implicit solution $y^2 = (\ln x)^2 + C$.
5. The logistic growth model is described by $\frac{dP}{dt} = P(1 - \frac{P}{K})$, where $K > 0$ is the carrying capacity. Find the general solution for $P(t)$.
   Answer: P(t) = \frac{K}{1 + A e^{-t}} The equation is separable: $\frac{dP}{P(1 - P/K)} = dt$. Rewrite the left side as $\frac{K dP}{P(K-P)} = dt$. Use partial fraction decomposition on $\frac{K}{P(K-P)} = \frac{1}{P} + \frac{1}{K-P}$. Integrating gives $\int (\frac{1}{P} + \frac{1}{K-P}) dP = \int dt$, which yields $\ln|P| - \ln|K-P| = t + C_1$. Combine the logarithms: $\ln|\frac{P}{K-P}| = t + C_1$. Exponentiate both sides: $|\frac{P}{K-P}| = e^{t+C_1} = e^{C_1}e^t$. Let $A_0 = \pm e^{C_1}$: $\frac{P}{K-P} = A_0 e^t$. Now solve for P: $P = A_0 e^t (K-P) = A_0 K e^t - A_0 P e^t$. $P(1 + A_0 e^t) = A_0 K e^t$. $P = \frac{A_0 K e^t}{1 + A_0 e^t}$. Divide numerator and denominator by $A_0 e^t$: $P = \frac{K}{1/A_0 e^{-t} + 1}$. Let $A = 1/A_0$. Then $P(t) = \frac{K}{1 + A e^{-t}}$.
6. Determine the general solution of the differential equation $\frac{dy}{dx} = x \sec(y)$.
   Answer: y = \arcsin($\frac{x^2}{2} + C$) The equation is separable. We can write it as $\frac{1}{\sec(y)} dy = x dx$, which simplifies to $\cos(y) dy = x dx$. Integrating both sides gives $\int \cos(y) dy = \int x dx$, which results in $\sin(y) = \frac{x^2}{2} + C$. To solve for $y$, we take the inverse sine of both sides, yielding $y = \arcsin(\frac{x^2}{2} + C)$.
7. What is the general solution to $\frac{dy}{dx} = \frac{1+y^2}{1+x^2}$?
   Answer: y = \frac{x+A}{1-Ax} Separating variables gives $\frac{1}{1+y^2} dy = \frac{1}{1+x^2} dx$. Integrating both sides gives $\arctan(y) = \arctan(x) + C$. To solve for $y$, we take the tangent of both sides: $y = \tan(\arctan(x) + C)$. Using the tangent addition formula $\tan(a+b) = \frac{\tan(a)+\tan(b)}{1-\tan(a)\tan(b)}$, we have $y = \frac{\tan(\arctan x) + \tan C}{1 - \tan(\arctan x)\tan C}$. Let $A = \tan(C)$, which is an arbitrary constant. This simplifies to $y = \frac{x+A}{1-Ax}$.
8. For the differential equation $\frac{dy}{dx} = \frac{\sin(x)}{y^2 + y}$, after factoring the denominator and using partial fractions, the separated form becomes:
   Answer: $\left(\frac{1}{y} - \frac{1}{y+1}\right)dy = \sin(x)dx$ When you encounter a separable differential equation like this one, your goal is to get all terms involving y on one side and all terms involving x on the other. The key challenge here is dealing with the denominator $y^2 + y$ using partial fraction decomposition. First, factor the denominator: $y^2 + y = y(y+1)$. Now you need to decompose $\frac{1}{y(y+1)}$ into partial fractions. Set up: $\frac{1}{y(y+1)} = \frac{A}{y} + \frac{B}{y+1}$ To find A and B, multiply both sides by $y(y+1)$: $1 = A(y+1) + By$ Setting $y = 0$: $1 = A(1) \Rightarrow A = 1$ Setting $y = -1$: $1 = B(-1) \Rightarrow B = -1$ Therefore: $\frac{1}{y(y+1)} = \frac{1}{y} - \frac{1}{y+1}$ The separated form becomes: $\left(\frac{1}{y} - \frac{1}{y+1}\right)dy = \sin(x)dx$ Choice A incorrectly puts y in the numerators instead of using the reciprocals from partial fractions. Choice B uses the correct format but has the wrong sign—it shows addition instead of subtraction. Choice C stops at the factored form without actually performing the partial fraction decomposition that the question specifically asks for. Study tip: When doing partial fractions with linear factors, remember that each factor $(ay+b)$ in the denominator contributes a term $\frac{A}{ay+b}$ to your decomposition. Always verify your A and B values by substituting convenient values of the variable.
9. Find the general solution for the differential equation $(1+x^2)y' - 2xy^2 = 0$.
   Answer: y = \frac{1}{C - \ln($1+x^2$)} First, rearrange the equation to $(1+x^2)\frac{dy}{dx} = 2xy^2$. Separate the variables: $\frac{1}{y^2} dy = \frac{2x}{1+x^2} dx$. Integrate both sides: $\int y^{-2} dy = \int \frac{2x}{1+x^2} dx$. The left side gives $-y^{-1}$. For the right side, use u-substitution with $u = 1+x^2$, $du = 2x dx$, so the integral is $\int \frac{1}{u} du = \ln|u| = \ln(1+x^2)$. The equation is $-\frac{1}{y} = \ln(1+x^2) + C_1$. Multiply by -1: $\frac{1}{y} = -\ln(1+x^2) - C_1$. Let $C = -C_1$. Then $\frac{1}{y} = C - \ln(1+x^2)$. Finally, take the reciprocal to find $y = \frac{1}{C - \ln(1+x^2)}$. Note that choice B is equivalent if C is replaced by -C, but the direct algebraic path leads to A.
10. An implicit form of the general solution to $\frac{dy}{dx} \sqrt{x} = y^3$ for $x > 0$ is given by:
    Answer: $y^2$ = \frac{1}{C - 4\sqrt{x}} Rearrange and separate variables: $y^{-3} dy = x^{-1/2} dx$. Integrate both sides: $\int y^{-3} dy = \int x^{-1/2} dx$. This gives $\frac{y^{-2}}{-2} = \frac{x^{1/2}}{1/2} + C_1$, which simplifies to $-\frac{1}{2y^2} = 2\sqrt{x} + C_1$. Multiply by -2: $\frac{1}{y^2} = -4\sqrt{x} - 2C_1$. Let $C = -2C_1$, a new arbitrary constant. Then $\frac{1}{y^2} = C - 4\sqrt{x}$. Inverting both sides gives $y^2 = \frac{1}{C - 4\sqrt{x}}$.
11. Find the general solution to the differential equation $(x^2+4)\frac{dy}{dx} = 1$.
    Answer: y = \frac{1}{2}\arctan($\frac{x}{2}$) + C The equation is separable. Rearrange it to $dy = \frac{1}{x^2+4} dx$. Integrate both sides: $\int dy = \int \frac{1}{x^2+4} dx$. The left side is $y$. The right side is a standard integral of the form $\int \frac{1}{a^2+u^2} du = \frac{1}{a}\arctan(\frac{u}{a})$. Here, $u=x$ and $a=2$. So, $\int \frac{1}{x^2+2^2} dx = \frac{1}{2}\arctan(\frac{x}{2})$. The general solution is $y = \frac{1}{2}\arctan(\frac{x}{2}) + C$.
12. Find the general solution to the differential equation $\frac{dy}{dx} = xy + 2y - x - 2$.
    Answer: y = 1 + A e^{\frac{x^2}{2} + 2x} The right side of the equation can be factored by grouping: $y(x+2) - 1(x+2) = (y-1)(x+2)$. The equation is $\frac{dy}{dx} = (y-1)(x+2)$. Separating variables gives $\frac{1}{y-1} dy = (x+2) dx$. Integrating both sides, $\int \frac{1}{y-1} dy = \int (x+2) dx$, yields $\ln|y-1| = \frac{x^2}{2} + 2x + C_1$. Exponentiating both sides gives $|y-1| = e^{\frac{x^2}{2} + 2x + C_1} = e^{C_1}e^{\frac{x^2}{2} + 2x}$. Let $A = \pm e^{C_1}$. Then $y-1 = A e^{\frac{x^2}{2} + 2x}$, and the general solution is $y = 1 + A e^{\frac{x^2}{2} + 2x}$.
13. Find the general solution to the differential equation $\frac{dy}{dx} = \frac{x \sin(x)}{e^y}$.
    Answer: y = \ln($\sin(x) - x\cos(x) + C$) Separate the variables to get $e^y dy = x \sin(x) dx$. Integrate both sides: $\int e^y dy = \int x \sin(x) dx$. The left integral is $e^y$. The right integral requires integration by parts. Let $u = x$ and $dv = \sin(x) dx$. Then $du = dx$ and $v = -\cos(x)$. The integral is $uv - \int v du = -x\cos(x) - \int (-\cos(x)) dx = -x\cos(x) + \int \cos(x) dx = -x\cos(x) + \sin(x) + C$. So, $e^y = \sin(x) - x\cos(x) + C$. Taking the natural logarithm gives $y = \ln(\sin(x) - x\cos(x) + C)$.
14. Find the general solution for the differential equation $\frac{dy}{dx} = \frac{x^2+1}{y^2}$.
    Answer: $y^3$ = $x^3$ + 3x + C When you encounter a differential equation like $\frac{dy}{dx} = \frac{x^2+1}{y^2}$, you're dealing with a separable differential equation. The key insight is recognizing that you can separate the variables by moving all terms involving $y$ to one side and all terms involving $x$ to the other. Starting with $\frac{dy}{dx} = \frac{x^2+1}{y^2}$, multiply both sides by $y^2$ and by $dx$ to get: $y^2 \, dy = (x^2+1) \, dx$ Now integrate both sides. The left side gives you $\int y^2 \, dy = \frac{y^3}{3}$, and the right side gives you $\int (x^2+1) \, dx = \frac{x^3}{3} + x$ This yields: $\frac{y^3}{3} = \frac{x^3}{3} + x + C$ Multiply everything by 3 to get the cleanest form: $y^3 = x^3 + 3x + C$ Answer choice A incorrectly uses $-\frac{1}{y}$ instead of $\frac{y^3}{3}$ when integrating $y^2$. Answer choice B has the wrong coefficient (3 instead of 1) in front of $y^3$ and missing the factor of 3 in front of $x$. Answer choice C fails to integrate the right side correctly, showing $2x$ instead of $\frac{x^3}{3} + x$. Study tip: For separable differential equations, always check your integration carefully and verify that your final answer makes sense by differentiating it back to the original equation.
15. Find the general solution to the differential equation $x^2 y' = y$, for $x > 0$.
    Answer: y = A $e^{-1/x}$ The equation can be written as $x^2 \frac{dy}{dx} = y$. Separating variables gives $\frac{1}{y} dy = \frac{1}{x^2} dx = x^{-2} dx$. Integrating both sides yields $\int \frac{1}{y} dy = \int x^{-2} dx$, which results in $\ln|y| = -x^{-1} + C = -\frac{1}{x} + C$. Exponentiating both sides to solve for $y$ gives $|y| = e^{-1/x + C} = e^C e^{-1/x}$. Let $A = \pm e^C$. The general solution is $y = A e^{-1/x}$.
16. The differential equation $\frac{dz}{dt} = t^2 z + t^2$ models a certain process. Find its general solution $z(t)$.
    Answer: z(t) = A $e^{t^3/3}$ - 1 First, factor the right-hand side: $\frac{dz}{dt} = t^2(z + 1)$. This is a separable differential equation. Separate the variables: $\frac{1}{z+1} dz = t^2 dt$. Integrate both sides: $\int \frac{1}{z+1} dz = \int t^2 dt$. This gives $\ln|z+1| = \frac{t^3}{3} + C_1$. Exponentiate both sides: $|z+1| = e^{t^3/3 + C_1} = e^{C_1}e^{t^3/3}$. Let $A = \pm e^{C_1}$. Then $z+1 = A e^{t^3/3}$. Finally, solve for $z$: $z(t) = A e^{t^3/3} - 1$.
17. The equation $\frac{dy}{dx} = \frac{x^2 - 1}{2y + 3}$ has a solution passing through $(1, -1)$. Which of the following is true about this solution?
    Answer: The solution exists and is unique in a neighborhood of $(1, -1)$ When you encounter a differential equation like this, you need to check whether a solution exists and is unique at the given point. This involves examining the equation's form and applying existence and uniqueness theorems. This differential equation can be written as $\frac{dy}{dx} = f(x,y)$ where $f(x,y) = \frac{x^2 - 1}{2y + 3}$. For the solution to exist and be unique near $(1, -1)$, you need both $f(x,y)$ and $\frac{\partial f}{\partial y}$ to be continuous in a neighborhood of that point. Let's check continuity at $(1, -1)$. First, the denominator: $2(-1) + 3 = 1 \neq 0$, so $f(x,y)$ is continuous near $(1, -1)$. The partial derivative $\frac{\partial f}{\partial y} = -\frac{x^2 - 1}{(2y + 3)^2}$ is also continuous there since the denominator doesn't equal zero. By the existence and uniqueness theorem, the solution exists and is unique in a neighborhood of $(1, -1)$. Option A incorrectly suggests the solution doesn't exist because $2(-1) + 3 = 1 \neq 0$, but this non-zero value actually ensures existence. Option B makes a calculation error, claiming $2(-1) + 3 = 0$, which is false. Option C correctly identifies that a solution exists but wrongly claims it's not unique—since both continuity conditions are met, uniqueness is guaranteed. Remember: for differential equations of the form $\frac{dy}{dx} = f(x,y)$, check if the denominator equals zero at your point. If not, and if the partial derivative exists, you typically have existence and uniqueness.
18. The differential equation $\frac{dy}{dx} = \frac{2x\cos(x^2)}{3y^2 + 1}$ can be solved by separation of variables. Which of the following represents the correct form after separation but before integration?
    Answer: $(3y^2 + 1)dy = 2x\cos(x^2)dx$ To separate variables, multiply both sides by $(3y^2 + 1)dx$: $dy = \frac{2x\cos(x^2)}{3y^2 + 1}dx$, then $(3y^2 + 1)dy = 2x\cos(x^2)dx$. Choice B incorrectly switches $dx$ and $dy$. Choice C fails to multiply the left side by $(3y^2 + 1)$. Choice D incorrectly inverts both sides.
19. A tank contains a solution where the rate of change of salt concentration $S$ with respect to time satisfies $\frac{dS}{dt} = \frac{S^2 - 4}{t^2 + 1}$. If $S(0) = 3$, what constraint must be considered when finding the domain of the solution?
    Answer: The solution may approach $S = 2$ or $S = -2$ asymptotically When separating variables: $\frac{dS}{S^2-4} = \frac{dt}{t^2+1}$. The separation is valid only when $S^2-4 \neq 0$, i.e., $S \neq \pm 2$. Since $S(0) = 3 > 2$, the solution might approach $S = 2$ as an asymptote, which would limit the domain. Choice A states when separation fails but doesn't address domain implications. Choice B is incorrect since $t^2+1 > 0$ always. Choice D ignores the asymptotic behavior near $S = 2$.
20. A radioactive substance decays according to $\frac{dN}{dt} = -\lambda \sqrt{N}$, where $N(t)$ is the amount at time $t$ and $\lambda > 0$. If $N(0) = N_0$, what is $N(t)$?
    Answer: $N(t) = \left(\sqrt{N_0} - \frac{\lambda t}{2}\right)^2$ Separating: $\frac{dN}{\sqrt{N}} = -\lambda dt$. Integrating: $\int N^{-1/2}dN = \int -\lambda dt$ gives $2\sqrt{N} = -\lambda t + C$. Using $N(0) = N_0$: $2\sqrt{N_0} = C$. So $2\sqrt{N} = -\lambda t + 2\sqrt{N_0}$, which gives $\sqrt{N} = \sqrt{N_0} - \frac{\lambda t}{2}$. Therefore $N(t) = \left(\sqrt{N_0} - \frac{\lambda t}{2}\right)^2$. Choice A is the solution to $\frac{dN}{dt} = -\lambda N$. Choice C has wrong sign. Choice D comes from incorrect integration.

Start this quiz