Finding General Solutions: Separation of Variables - AP Calculus AB

A quantity $y$ changes with $x$ by $\dfrac{dy}{dx}=\dfrac{3}{x}y$ . What is the general solution?
Answer: $y=Cx^{3}$ This differential equation $\frac{dy}{dx} = \frac{3}{x}y$ is perfectly suited for separation of variables. We can rearrange it as $\frac{dy}{y} = \frac{3}{x}dx$ , placing all y terms on the left and all x terms on the right. Integrating both sides gives $\ln|y| = 3\ln|x| + C_1$ , which simplifies to $\ln|y| = \ln|x^3| + C_1$ . Exponentiating both sides yields $y = e^{\ln|x^3| + C_1} = e^{C_1} \cdot |x|^3$ . Since we typically work with $x > 0$ in these contexts and $e^{C_1}$ is an arbitrary constant, we write the general solution as $y = Cx^3$ . Choice A incorrectly integrates as if y were a constant rather than the dependent variable. When you see a differential equation where the derivative equals a product with one factor containing only x and another containing only y, separation of variables is your go-to method.
A function satisfies $\dfrac{dy}{dx}=2xy$ . What is the general solution for $y(x)$ ?
Answer: $y=Ce^{x^{2}}$ This problem requires solving a differential equation using separation of variables. For dy/dx = 2 x y, separate as dy/y = 2 x dx, assuming y ≠ 0. Integrate: ln|y| = $x^2$ + C. Exponentiate to y = C $e^{x^2}$ . The distractor y = $e^{x^2}$ + C fails by not using a multiplicative constant. To recognize separable equations, look for the form dy/dx = f(x) g(y), which allows rewriting as (1/g(y)) dy = f(x) dx for integration.
A model satisfies $\frac{dy}{dx}=\frac{3x}{1+x^2}y$ . What is the general solution for $y(x)$ ?
Answer: $y=C(1+x^2)^{3/2}$ This differential equation uses separation of variables with a rational function. Starting with $\frac{dy}{dx}=\frac{3x}{1+x^2}y$ , we separate to get $\frac{dy}{y}=\frac{3x}{1+x^2}dx$ . To integrate the right side, note that $\frac{d}{dx}[\ln(1+x^2)]=\frac{2x}{1+x^2}$ , so $\int\frac{3x}{1+x^2}dx=\frac{3}{2}\ln(1+x^2)$ . This gives $\ln|y|=\frac{3}{2}\ln(1+x^2)+C_1=\ln(1+x^2)^{3/2}+C_1$ . Exponentiating yields $|y|=e^{C_1}(1+x^2)^{3/2}$ , which becomes $y=C(1+x^2)^{3/2}$ where $C=\pm e^{C_1}$ . Choice C omits the arbitrary constant. When integrating rational functions of the form $\frac{kx}{1+x^2}$ , recognize this as a logarithmic derivative pattern.
A cooling object satisfies $\frac{dT}{dt}=4(T-10)$ . What is the general solution for $T(t)$ ?
Answer: $T(t)=10+Ce^{4t}$ This Newton's Law of Cooling equation dT/dt = 4(T-10) requires separation of variables after recognizing the shifted form. We can rewrite this as dT/dt = 4T - 40, and to separate variables, we need dT/(T-10) = 4dt. Integrating both sides gives ln|T-10| = 4t + C₁. Exponentiating yields |T-10| = $e^{4t+C₁}$ = $Ce^{4t}$ , where C = ± $e^C$ ₁. Solving for T gives T = 10 + $Ce^{4t}$ , which is the general solution. Choice B incorrectly places the 10 as -10 outside the exponential term, which would give dT/dt = $4Ce^{4t}$ ≠ 4(T-10). When dealing with shifted exponential models like dT/dt = k(T-a), the solution is always T = a + $Ce^{kt}$ , not $Ce^{kt}$ - a.
A radioactive sample satisfies $\frac{dA}{dt}=-\frac{1}{5}A$ . What is the general solution for $A(t)$ ?
Answer: $A(t)=Ce^{-t/5}$ This radioactive decay equation dA/dt = -A/5 is solved using separation of variables. Separating the variables gives us dA/A = -dt/5, dividing both sides by A and multiplying by dt. Integrating both sides yields ln|A| = -t/5 + C₁, where C₁ is the constant of integration. Exponentiating both sides gives |A| = $e^{-t/5+C₁}$ = $e^C$ ₁ · $e^{-t/5}$ , which simplifies to A = $Ce^{-t/5}$ where C = ± $e^C$ ₁ is an arbitrary constant. Choice E incorrectly has a positive exponent $e^{t/5}$ , which would represent exponential growth rather than the decay indicated by the negative coefficient in the differential equation. When the coefficient of the dependent variable is negative in dA/dt = kA, the solution always involves $e^{kt}$ with k negative.
A function satisfies $\dfrac{dy}{dx}=\dfrac{6x^{2}}{y}$ . What is the general solution?
Answer: $\dfrac{1}{2}y^{2}=2x^{3}+C$ This differential equation $\frac{dy}{dx} = \frac{6x^2}{y}$ requires separation of variables. We multiply both sides by y and dx to get $y dy = 6x^2 dx$ , separating the variables. Integrating both sides gives us $\int y dy = \int 6x^2 dx$ , which evaluates to $\frac{y^2}{2} = 6 \cdot \frac{x^3}{3} + C = 2x^3 + C$ . This is the general solution in implicit form. Choice A incorrectly computes the integral of $6x^2$ as $6x^3$ instead of $2x^3$ , forgetting to divide by 3. When separating variables leads to polynomial integrands on both sides, carefully apply the power rule and don't forget the constants that arise from integration.
A culture grows so that $\frac{dP}{dt}=3P$ . What is the general solution for $P$ as a function of $t$ ?
Answer: $P(t)=Ce^{3t}$ This problem requires separation of variables to solve the differential equation $\frac{dP}{dt}=3P$ . To separate variables, we divide both sides by $P$ and multiply by $dt$ , giving us $\frac{dP}{P}=3dt$ . Integrating both sides yields $\ln|P|=3t+C_1$ , where $C_1$ is an arbitrary constant. Exponentiating both sides gives $|P|=e^{3t+C_1}=e^{C_1}e^{3t}$ , which we can write as $P=Ce^{3t}$ where $C=\pm e^{C_1}$ is an arbitrary constant. Choice D incorrectly stops at the logarithmic form without exponentiating to solve for $P$ . When you see a differential equation of the form $\frac{dy}{dt}=ky$ , recognize it as exponential growth/decay and expect a solution $y=Ce^{kt}$ .
A spring model gives $\frac{dy}{dx}=\frac{y}{1+x}$ for $x>-1$ . What is the general solution?
Answer: $y(x)=C(1+x)$ This differential equation dy/dx = y/(1+x) is solved using separation of variables. Separating gives us dy/y = dx/(1+x), with y terms on the left and x terms on the right. Integrating both sides yields ln|y| = ln|1+x| + C₁, where we recognize that the integral of 1/(1+x) is ln|1+x|. Using logarithm properties, we can write ln|y| = ln|1+x| + ln|C| = ln|C(1+x)|, where C = $e^C$ ₁. Exponentiating gives y = C(1+x), which is the general solution. Choice B incorrectly suggests y = ln(1+x) + C, but this would give dy/dx = 1/(1+x), not y/(1+x) as required. Remember that when integrating gives matching logarithmic forms, the solution often involves the arguments of those logarithms in a product.
A function satisfies $\dfrac{dy}{dx}=x y$ . What is the general solution for $y$ ?
Answer: $y=Ce^{x^{2}/2}$ This problem requires solving a differential equation using separation of variables. For dy/dx = x y, separate as dy/y = x dx, assuming y ≠ 0. Integrate: ln|y| = ( $x^2$ )/2 + C. Exponentiate to y = C $e^{x^2/2}$ . The distractor y = $e^{x^2/2}$ + C fails by adding C outside, not as a multiplier. To recognize separable equations, look for the form dy/dx = f(x) g(y), which allows rewriting as (1/g(y)) dy = f(x) dx for integration.
A variable satisfies $\dfrac{dy}{dx}=\dfrac{2}{x}y$ for $x\ne 0$ . What is the general solution?
Answer: $y=Cx^{2}$ This problem requires solving a differential equation using separation of variables. For dy/dx = (2/x) y with x ≠ 0, separate as dy/y = (2/x) dx, assuming y ≠ 0. Integrate: ln|y| = 2 ln|x| + C. Simplify to y = C $x^2$ , absorbing constants. The distractor y = 2 ln|x| + C fails by not exponentiating properly, yielding a logarithmic form instead of power. To recognize separable equations, look for the form dy/dx = f(x) g(y), which allows rewriting as (1/g(y)) dy = f(x) dx for integration.
A quantity satisfies $\dfrac{dy}{dx}=\dfrac{y}{1+x^{2}}$ . What is the general solution?
Answer: $y=Ce^{\arctan(x)}$ This problem requires solving a differential equation using separation of variables. For dy/dx = y / (1 + $x^2$ ), separate as dy/y = dx/(1 + $x^2$ ), assuming y ≠ 0. Integrate: ln|y| = arctan(x) + C. Exponentiate to y = C $e^{arctan(x)}$ . The distractor y = $e^{arctan(x)}$ + C fails by not incorporating the multiplicative constant from exponentiation. To recognize separable equations, look for the form dy/dx = f(x) g(y), which allows rewriting as (1/g(y)) dy = f(x) dx for integration.
A function satisfies $\dfrac{dy}{dx}=\dfrac{y}{2x}$ . What is the general solution for $x\ne 0$ ?
Answer: $y=Cx^{1/2}$ This problem requires solving a differential equation using separation of variables. For $dy/dx = y/(2x)$ with x ≠ 0, separate as $dy/y = (1/2) dx/x$ , assuming y ≠ 0. Integrate: $ln|y| = (1/2) ln|x| + C$ . Simplify to $y = C x^{1/2}$ . The distractor $y = (1/2) ln|x| + C$ fails by not completing the exponentiation step. To recognize separable equations, look for the form $dy/dx = f(x) g(y)$ , which allows rewriting as $(1/g(y)) dy = f(x) dx$ for integration.
A population satisfies $\dfrac{dP}{dt}=\dfrac{P}{4}$ . What is the general solution for $P(t)$ ?
Answer: $P=Ce^{t/4}$ This problem requires solving a differential equation using separation of variables. Given dP/dt = P/4, separate as dP/P = (1/4) dt, assuming P ≠ 0. Integrate to get ln|P| = (t/4) + C. Exponentiate for P = $Ce^{t/4}$ . The choice P = $e^{t/4}$ + C is incorrect because it adds a constant outside the exponential, unlike the proper form from integration. To recognize separable equations, look for the form dy/dx = f(x) g(y), which allows rewriting as (1/g(y)) dy = f(x) dx for integration.
A quantity $y$ satisfies $\dfrac{dy}{dx}=y^{2}$ . What is the general solution for $y(x)$ ?
Answer: $y=-\dfrac{1}{x+C}$ This differential equation $\frac{dy}{dx} = y^2$ is a classic separation of variables problem. We rearrange to get $\frac{dy}{y^2} = dx$ , separating y terms and x terms. Integrating both sides gives us $\int y^{-2} dy = \int dx$ , which evaluates to $-y^{-1} = x + C$ , or $-\frac{1}{y} = x + C$ . Solving for y, we multiply both sides by -1 and take the reciprocal: $y = -\frac{1}{x + C}$ . Choice A has the wrong sign, while choice C leaves the solution in implicit form rather than solving for y. When you encounter $\frac{dy}{dx} = y^n$ where n ≠ 1, separation of variables followed by the power rule for integration will lead you to the general solution.
A bacteria count satisfies $\frac{dB}{dt}=-\frac{3}{2}B$ . What is the general solution for $B(t)$ ?
Answer: $B(t)=Ce^{-3t/2}$ To solve $\frac{dB}{dt}=-\frac{3}{2}B$ using separation of variables, we divide by $B$ and multiply by $dt$ to get $\frac{dB}{B}=-\frac{3}{2}dt$ . Integrating gives $\ln|B|=-\frac{3}{2}t+C_1$ . Exponentiating yields $|B|=e^{-3t/2+C_1}=e^{C_1}e^{-3t/2}$ , which simplifies to $B=Ce^{-3t/2}$ where $C=\pm e^{C_1}$ . Choice B incorrectly uses a positive exponent, ignoring the negative sign in the original differential equation. The negative coefficient indicates bacterial decay, which must result in a negative exponent in the exponential solution.
A function satisfies $\dfrac{dy}{dx}=\dfrac{2}{y}$ . What is the general solution relating $x$ and $y$ ?
Answer: $\dfrac{1}{2}y^{2}=2x+C$ This differential equation $\frac{dy}{dx} = \frac{2}{y}$ is ideal for separation of variables. We multiply both sides by y and dx to get $y dy = 2 dx$ , successfully separating the variables. Integrating both sides gives us $\int y dy = \int 2 dx$ , which evaluates to $\frac{y^2}{2} = 2x + C$ . This is already in its simplest form as the general solution. Choice C incorrectly omits the factor of $\frac{1}{2}$ that comes from integrating y, while choice D would result from incorrectly treating this as $\frac{dy}{dx} = 2y$ instead. The key insight is that when dy/dx equals a function of y in the denominator, multiply both sides by y before integrating to properly separate variables.
A motion model satisfies $\frac{dv}{dt}=\frac{2}{t}v$ for $t>0$ . What is the general solution for $v(t)$ ?
Answer: $v(t)=Ct^2$ This differential equation uses separation of variables with a rational function coefficient. From $\frac{dv}{dt}=\frac{2}{t}v$ , we separate to get $\frac{dv}{v}=\frac{2}{t}dt$ . Integrating both sides gives $\ln|v|=2\ln|t|+C_1=\ln|t|^2+C_1$ , using the fact that $\int\frac{2}{t}dt=2\ln|t|$ . Exponentiating yields $|v|=e^{\ln|t|^2+C_1}=e^{C_1}|t|^2$ , which becomes $v=Ct^2$ where $C=\pm e^{C_1}$ for $t>0$ . Choice A incorrectly integrates $\frac{2}{t}$ as if it were multiplying v directly. When you encounter $\frac{dy}{dx}=\frac{f(x)}{x}y$ , the solution often involves powers of x rather than exponentials.
A function satisfies $\dfrac{dy}{dx}=\dfrac{2y}{\sqrt{x}}$ for $x>0$ . What is the general solution?
Answer: $y=Ce^{4\sqrt{x}}$ This problem requires solving a differential equation using separation of variables. Given dy/dx = 2 y / sqrt(x) for x > 0, separate as dy/y = 2 dx / sqrt(x), assuming y ≠ 0. Integrate to ln|y| = 4 sqrt(x) + C. Exponentiate for y = C $e^{4 sqrt(x)}$ . The choice y = $e^{4 sqrt(x)}$ + C is incorrect as it doesn't multiply by C. To recognize separable equations, look for the form dy/dx = f(x) g(y), which allows rewriting as (1/g(y)) dy = f(x) dx for integration.
A function $y(x)$ satisfies $\dfrac{dy}{dx}=\dfrac{x}{y}$ . What is the general solution?
Answer: $y^{2}=x^{2}+C$ This differential equation $\frac{dy}{dx} = \frac{x}{y}$ requires separation of variables to solve. We multiply both sides by y and dx to get $y dy = x dx$ , successfully separating the variables. Integrating both sides gives us $\int y dy = \int x dx$ , which evaluates to $\frac{y^2}{2} = \frac{x^2}{2} + C_1$ . Multiplying through by 2, we get $y^2 = x^2 + 2C_1$ , and since $2C_1$ is just another arbitrary constant, we write the general solution as $y^2 = x^2 + C$ . Choice B would incorrectly suggest that y and x are linearly related, missing the quadratic nature revealed by integration. When the derivative is a ratio with x in the numerator and y in the denominator, cross-multiply before integrating to properly separate variables.
A model satisfies $\frac{dy}{dx}=\frac{y}{1+x^2}$ . What is the general solution for $y(x)$ ?
Answer: $y=Ce^{\arctan(x)}$ This differential equation uses separation of variables with a trigonometric integral. From $\frac{dy}{dx}=\frac{y}{1+x^2}$ , we separate to get $\frac{dy}{y}=\frac{dx}{1+x^2}$ . Integrating both sides gives $\ln|y|=\arctan(x)+C_1$ , since $\int\frac{1}{1+x^2}dx=\arctan(x)$ . Exponentiating yields $|y|=e^{\arctan(x)+C_1}=e^{C_1}e^{\arctan(x)}$ , which becomes $y=Ce^{\arctan(x)}$ where $C=\pm e^{C_1}$ . Choice C omits the arbitrary constant C, giving only a particular solution rather than the general solution. Always include the arbitrary constant when finding general solutions to differential equations.

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