General Solutions: Separation of Variables - Calculus 1

1. Solve the differential equation $y' = y^2 \sin(x)$.
   Answer: $y = \frac{1}{\cos(x)+C}$ This is a separable differential equation, which means you can separate the variables and integrate both sides. When you see an equation of the form $y' = f(x)g(y)$, your strategy is to rewrite it as $\frac{dy}{dx} = f(x)g(y)$ and then separate variables. Starting with $y' = y^2 \sin(x)$, rewrite as $\frac{dy}{dx} = y^2 \sin(x)$. Separate variables by dividing both sides by $y^2$ and multiplying by $dx$: $\frac{dy}{y^2} = \sin(x) dx$. Now integrate both sides: $\int y^{-2} dy = \int \sin(x) dx$. The left side gives $\frac{y^{-1}}{-1} = -\frac{1}{y}$, and the right side gives $-\cos(x)$. So you have $-\frac{1}{y} = -\cos(x) + C$. Solving for $y$: $\frac{1}{y} = \cos(x) - C$, which means $y = \frac{1}{\cos(x) - C}$. Since $C$ is an arbitrary constant, $-C$ is also arbitrary, so we can write this as $y = \frac{1}{\cos(x) + C}$. Choice A gives $y = \frac{-1}{\cos(x)} + C$, which has the wrong form entirely—it's not even a solution to our separated equation. Choice B, $y = -\cos(x) + C$, would give $y' = \sin(x)$, not $y^2\sin(x)$. Choice C, $y = \ln(\sec(x)) + C$, differentiates to $y' = \tan(x)$, again missing the $y^2$ term. For separable differential equations, always check your algebra when solving for the dependent variable—sign errors and reciprocal mistakes are common pitfalls.
2. What is the general solution to the differential equation $\frac{dy}{dx} = e^{x+y}$?
   Answer: $y = -\ln(C - e^x)$ First, use the property of exponents to rewrite the equation: $\frac{dy}{dx} = e^x e^y$. Separate the variables by dividing by $e^y$: $\frac{1}{e^y} dy = e^x dx$, which is $e^{-y} dy = e^x dx$. Integrate both sides: $\int e^{-y} dy = \int e^x dx$. This gives $-e^{-y} = e^x + C_1$. To solve for $y$, first multiply by -1: $e^{-y} = -e^x - C_1$. Let $C = -C_1$. Then $e^{-y} = C - e^x$. Take the natural logarithm of both sides: $-y = \ln(C - e^x)$. Finally, multiply by -1: $y = -\ln(C - e^x)$. Distractor B results from a sign error when integrating $e^{-y}$. Distractors C and D result from fundamental misunderstandings of how to solve this type of differential equation.
3. Which of the following represents the general solution to the differential equation $y' = e^{2x+y}$?
   Answer: $y = -\ln(-\frac{1}{2}e^{2x} + C)$ First, use exponent rules to write $y' = e^{2x}e^y$. Separate the variables: $e^{-y}dy = e^{2x}dx$. Integrate both sides: $\int e^{-y}dy = \int e^{2x}dx$, which gives $-e^{-y} = \frac{1}{2}e^{2x} + K$. To solve for y, first multiply by -1: $e^{-y} = -\frac{1}{2}e^{2x} - K$. Let $C = -K$, so $e^{-y} = C - \frac{1}{2}e^{2x}$. Now, take the natural logarithm of both sides: $-y = \ln(C - \frac{1}{2}e^{2x})$. Finally, $y = -\ln(C - \frac{1}{2}e^{2x})$, which is equivalent to choice B.
4. Given $\frac{dy}{dx} = \sqrt{y} \sin(x)$ for $y > 0$, find the general solution for $y$.
   Answer: $y = (K - \frac{1}{2}\cos(x))^2$ Separate the variables: $\frac{1}{\sqrt{y}} dy = \sin(x) dx$, or $y^{-1/2} dy = \sin(x) dx$. Integrate both sides: $\int y^{-1/2} dy = \int \sin(x) dx$. Using the power rule for the left side and the standard integral for the right side, we get: $2y^{1/2} = -\cos(x) + C_1$. Now, solve for $y$. First, divide by 2: $y^{1/2} = -\frac{1}{2}\cos(x) + \frac{C_1}{2}$. Let $K = \frac{C_1}{2}$ be the new arbitrary constant. The equation is $\sqrt{y} = K - \frac{1}{2}\cos(x)$. Finally, square both sides to get the explicit solution: $y = (K - \frac{1}{2}\cos(x))^2$. Distractor A has a sign error on the integral of $\sin(x)$. Distractor B incorrectly handles the constant 2 from the integration. Distractor D incorrectly multiplies by 2 instead of dividing.
5. Find the general solution for $y$ given the differential equation $\frac{dy}{dx} = \frac{y^2+1}{x^2}$.
   Answer: $y = \tan(C - \frac{1}{x})$ First, separate the variables: $\frac{1}{y^2+1} dy = \frac{1}{x^2} dx$. Next, integrate both sides: $\int \frac{1}{y^2+1} dy = \int x^{-2} dx$. The integral of the left side is $\arctan(y)$. The integral of the right side is $-x^{-1} + C$, or $C - \frac{1}{x}$. So, we have the equation $\arctan(y) = C - \frac{1}{x}$. To solve for $y$, we take the tangent of both sides: $y = \tan(C - \frac{1}{x})$. Distractor B incorrectly distributes the tangent function. Distractor C forgets to apply the tangent function to solve for $y$. Distractor D uses an incorrect antiderivative for $1/x^2$.
6. What is the implicit general solution to the differential equation $(1+y^2)dx + (1+x^2)dy = 0$?
   Answer: $\arctan(y) + \arctan(x) = C$ Rearrange the equation to $(1+x^2)dy = -(1+y^2)dx$. Separate variables: $\frac{dy}{1+y^2} = -\frac{dx}{1+x^2}$. Integrate both sides: $\int \frac{dy}{1+y^2} = -\int \frac{dx}{1+x^2}$. This gives $\arctan(y) = -\arctan(x) + C$. Rearranging gives the implicit solution $\arctan(y) + \arctan(x) = C$.
7. What is the general solution to $x \frac{dy}{dx} = y \ln(x)$?
   Answer: $y = C e^{\frac{1}{2}(\ln x)^2}$ Separate the variables: $\frac{1}{y}dy = \frac{\ln(x)}{x}dx$. Integrate both sides. For the right side, use u-substitution with $u = \ln(x)$, so $du = \frac{1}{x}dx$. The integral becomes $\int \frac{1}{y}dy = \int u du$. This gives $\ln|y| = \frac{1}{2}u^2 + K = \frac{1}{2}(\ln x)^2 + K$. To solve for y, exponentiate both sides: $|y| = e^{\frac{1}{2}(\ln x)^2 + K} = e^K e^{\frac{1}{2}(\ln x)^2}$. Let $C = \pm e^K$. The solution is $y = C e^{\frac{1}{2}(\ln x)^2}$.
8. Find the general solution to $y' = (y+1)^2$.
   Answer: $y = \frac{1}{C-x} - 1$ When you encounter a differential equation like $y' = (y+1)^2$, you're looking at a separable differential equation. The key is to separate the variables so that all terms involving $y$ are on one side and all terms involving $x$ are on the other. Starting with $\frac{dy}{dx} = (y+1)^2$, you can rewrite this as $\frac{dy}{(y+1)^2} = dx$. Now integrate both sides. The left side requires the substitution $u = y+1$, giving us $\int (y+1)^{-2} dy = \int dx$, which becomes $-\frac{1}{y+1} = x + C_1$. Solving for $y$, multiply both sides by $-1$: $\frac{1}{y+1} = -x - C_1$. Let $C = -C_1$ to simplify: $\frac{1}{y+1} = C - x$. Therefore, $y+1 = \frac{1}{C-x}$, and finally $y = \frac{1}{C-x} - 1$. This matches answer choice D. Answer A represents the solution to $y' = 1 + y^2$, not $(y+1)^2$. Answer B incorrectly suggests taking the natural logarithm of the right side, which isn't the proper integration technique for this separable equation. Answer C has the wrong exponent - it would result from integrating incorrectly or confusing the separation step. Remember: for separable equations of the form $y' = f(y)$, always separate variables first, then integrate. Watch out for the negative sign when integrating $\frac{1}{(y+1)^2}$ - it's a common source of errors.
9. The general solution to the differential equation $y' = \frac{y}{x \ln x}$ can be written in the form $y = f(x)$. What is $f(x)$?
   Answer: $f(x) = A \ln(x)$ First, write the equation as $\frac{dy}{dx} = \frac{y}{x \ln x}$ and separate the variables: $\frac{1}{y} dy = \frac{1}{x \ln x} dx$. Next, integrate both sides: $\int \frac{1}{y} dy = \int \frac{1}{x \ln x} dx$. The left integral is $\ln|y|$. For the right integral, use u-substitution with $u = \ln x$, so $du = \frac{1}{x} dx$. The integral becomes $\int \frac{1}{u} du = \ln|u| = \ln|\ln x|$. So, we have $\ln|y| = \ln|\ln x| + C$. To solve for $y$, exponentiate both sides: $|y| = e^{\ln|\ln x| + C} = e^{\ln|\ln x|} \cdot e^C = |\ln x| \cdot e^C$. Let $A = \pm e^C$. Then $y = A \ln(x)$.
10. The general solution to the differential equation $\frac{dy}{dx} = \frac{2x+1}{2y}$ is a family of curves. Which equation best describes this family?
    Answer: $y^2 = x^2 + x + C$ Separate the variables: $2y\,dy = (2x+1)\,dx$. Integrate both sides: $\int 2y\,dy = \int (2x+1)\,dx$. This gives $y^2 = x^2 + x + C$. This is the implicit form of the solution. Distractor A incorrectly places the constant outside the square root. Distractor C has an error in the integration of 2y. Distractor D incorrectly takes the square root of the terms individually.
11. The rate of change of a quantity $y$ with respect to $x$ is given by $\frac{dy}{dx} = \frac{xy}{x^2+1}$. Find the general solution for $y$.
    Answer: $y = C\sqrt{x^2+1}$ Separate the variables: $\frac{1}{y}dy = \frac{x}{x^2+1}dx$. Integrate both sides. For the right side, use u-substitution with $u=x^2+1$, so $du=2x\,dx$ and $x\,dx = du/2$. The integral becomes $\int \frac{1}{y}dy = \int \frac{1}{u} \frac{du}{2}$. This gives $\ln|y| = \frac{1}{2}\ln|u| + K = \frac{1}{2}\ln(x^2+1) + K$. Using log properties, $\ln|y| = \ln((x^2+1)^{1/2}) + K = \ln(\sqrt{x^2+1}) + K$. Exponentiating gives $|y| = e^K e^{\ln(\sqrt{x^2+1})} = e^K \sqrt{x^2+1}$. Let $C = \pm e^K$, so $y = C\sqrt{x^2+1}$.
12. The general solution to the differential equation $\frac{dy}{dx} = -4x(y-1)$ is of the form $y = 1 + f(x, C)$. What is $f(x, C)$?
    Answer: $Ce^{-2x^2}$ When you encounter a differential equation like $\frac{dy}{dx} = -4x(y-1)$, recognize this as a separable differential equation. The key insight is that you can separate variables by moving all terms involving $y$ to one side and all terms involving $x$ to the other. Starting with $\frac{dy}{dx} = -4x(y-1)$, separate variables: $\frac{dy}{y-1} = -4x \, dx$ Now integrate both sides. The left side gives $\ln|y-1|$, and the right side gives $-2x^2 + C_1$. So: $\ln|y-1| = -2x^2 + C_1$ Exponentiating both sides: $|y-1| = e^{-2x^2 + C_1} = e^{C_1} \cdot e^{-2x^2}$ Since $e^{C_1}$ is just another constant, call it $C$: $y-1 = Ce^{-2x^2}$ Therefore: $y = 1 + Ce^{-2x^2}$ This matches the form $y = 1 + f(x,C)$ where $f(x,C) = Ce^{-2x^2}$, which is choice D. Choice A incorrectly includes $(y-1)$ in the function $f$, but $f$ should only depend on $x$ and $C$. Choice B has the wrong exponent—it uses $-4x$ instead of $-2x^2$, missing the integration of $x$. Choice C adds $C$ to the exponential instead of multiplying by it, which doesn't arise from the integration process. Study tip: For separable differential equations, always check your integration carefully—especially when integrating polynomials like $x$, where $\int -4x \, dx = -2x^2$, not $-4x$.
13. Find the general solution for $y(x)$ given that $y' = x\sqrt{1-y^2}$.
    Answer: $y(x) = \sin(\frac{x^2}{2} + C)$ Separate the variables: $\frac{1}{\sqrt{1-y^2}}dy = x\,dx$. Integrate both sides: $\int \frac{1}{\sqrt{1-y^2}}dy = \int x\,dx$. The left integral is $\arcsin(y)$ and the right is $\frac{x^2}{2} + C$. So, $\arcsin(y) = \frac{x^2}{2} + C$. To solve for $y$, take the sine of both sides: $y = \sin(\frac{x^2}{2} + C)$. Distractor A incorrectly adds C after applying the sine function. Distractors B and D use incorrect antiderivatives.
14. Find the general solution to the differential equation $(1+x^2) dy - 2xy dx = 0$.
    Answer: $y = A(1+x^2)$ First, rearrange the equation to prepare for separation: $(1+x^2) dy = 2xy dx$. Now, separate the variables by moving $y$ terms to the left and $x$ terms to the right: $\frac{1}{y} dy = \frac{2x}{1+x^2} dx$. Integrate both sides: $\int \frac{1}{y} dy = \int \frac{2x}{1+x^2} dx$. The left integral is $\ln|y|$. For the right integral, use u-substitution with $u = 1+x^2$, so $du = 2x dx$. The integral becomes $\int \frac{1}{u} du = \ln|u| = \ln(1+x^2)$ (absolute value is not needed since $1+x^2 > 0$). So, $\ln|y| = \ln(1+x^2) + C$. Exponentiate both sides: $|y| = e^{\ln(1+x^2)+C} = e^{\ln(1+x^2)}e^C = (1+x^2)e^C$. Let $A = \pm e^C$. The solution is $y = A(1+x^2)$. Distractor A represents the common error of treating $e^{a+b}$ as $e^a+e^b$. Distractor C results from incorrectly integrating $\frac{2x}{1+x^2}$ as if it were $\frac{1}{1+x^2}$.
15. Find the general solution to the differential equation $\frac{dy}{dx} = xy + 3x - 2y - 6$.
    Answer: $y = Ce^{\frac{1}{2}x^2 - 2x} - 3$ The right side of the equation can be factored by grouping: $xy + 3x - 2y - 6 = x(y+3) - 2(y+3) = (x-2)(y+3)$. The equation becomes $\frac{dy}{dx} = (x-2)(y+3)$. Separating variables gives $\frac{1}{y+3}dy = (x-2)dx$. Integrating both sides yields $\ln|y+3| = \frac{1}{2}x^2 - 2x + K$. Exponentiating both sides gives $|y+3| = e^{\frac{1}{2}x^2 - 2x + K} = e^K e^{\frac{1}{2}x^2 - 2x}$. Let $C = \pm e^K$. Then $y+3 = Ce^{\frac{1}{2}x^2 - 2x}$, so $y = Ce^{\frac{1}{2}x^2 - 2x} - 3$.
16. Find the general solution, in implicit form, to the differential equation $xy \frac{dy}{dx} = y+1$.
    Answer: $y - \ln|y+1| = \ln|x| + C$ To separate the variables, move all $y$ terms to the left and all $x$ terms to the right: $\frac{y}{y+1} dy = \frac{1}{x} dx$. To integrate the left side, $\int \frac{y}{y+1} dy$, we can rewrite the integrand as $\frac{y+1-1}{y+1} = 1 - \frac{1}{y+1}$. Now integrate both sides: $\int (1 - \frac{1}{y+1}) dy = \int \frac{1}{x} dx$. This yields $y - \ln|y+1| = \ln|x| + C$. This is the implicit general solution. Distractor B has a sign error in the integration of the $y$ term. Distractor C results from an incorrect attempt to integrate $\frac{y}{y+1}$. Distractor D uses the wrong antiderivative for $\frac{1}{x}$.
17. Find the general solution to the differential equation $y' = y + e^x y$.
    Answer: $y = C e^{x+e^x}$ First, factor the right side: $y' = y(1+e^x)$. Separate the variables: $\frac{1}{y}dy = (1+e^x)dx$. Integrate both sides: $\int \frac{1}{y}dy = \int (1+e^x)dx$. This gives $\ln|y| = x + e^x + K$. Exponentiate both sides: $|y| = e^{x+e^x+K} = e^K e^{x+e^x}$. Let $C = \pm e^K$. The general solution is $y = C e^{x+e^x}$.
18. Determine the general solution of the differential equation $\frac{dy}{dx} - x^2y = 0$.
    Answer: $y = C e^{x^3/3}$ First, rearrange the equation to $\frac{dy}{dx} = x^2y$. Separate the variables: $\frac{1}{y}dy = x^2 dx$. Integrate both sides: $\int \frac{1}{y}dy = \int x^2 dx$, which results in $\ln|y| = \frac{x^3}{3} + K$. Exponentiating both sides gives $|y| = e^{x^3/3 + K} = e^K e^{x^3/3}$. Let $C = \pm e^K$. The general solution is $y = C e^{x^3/3}$.
19. The general solution of $\frac{dy}{dx} = \frac{1}{y \cdot x \ln x}$ is $y^2 = f(x, C)$. What is $f(x, C)$?
    Answer: $2\ln|\ln x| + C$ Separate the variables: $y\,dy = \frac{1}{x \ln x}dx$. Integrate both sides. The left side is $\frac{y^2}{2}$. For the right side, use u-substitution with $u = \ln x$, so $du = \frac{1}{x}dx$. The integral is $\int \frac{1}{u}du = \ln|u| = \ln|\ln x|$. So, we have $\frac{y^2}{2} = \ln|\ln x| + K$. Multiply by 2: $y^2 = 2\ln|\ln x| + 2K$. Let $C = 2K$. Thus, $y^2 = 2\ln|\ln x| + C$.
20. Which expression represents the general solution to the differential equation $\frac{dy}{dx} = \frac{\cot(y)}{\cot(x)}$?
    Answer: $\sec(y) = C\sec(x)$ The equation can be written as $\frac{dy}{dx} = \cot(y)\tan(x)$. Separating variables gives $\frac{1}{\cot(y)}dy = \tan(x)dx$, which is $\tan(y)dy = \tan(x)dx$. Integrating both sides: $\int \tan(y)dy = \int \tan(x)dx$. This yields $\ln|\sec(y)| = \ln|\sec(x)| + K$. Exponentiate both sides: $e^{\ln|\sec(y)|} = e^{\ln|\sec(x)| + K}$, which simplifies to $|\sec(y)| = e^K|\sec(x)|$. Let $C = \pm e^K$. This gives $\sec(y) = C\sec(x)$.

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