Finding General Solutions: Separation of Variables - AP Calculus BC

A function $y(x)$ satisfies $\frac{dy}{dx}=(x^2+1)e^{y}$ . What is the general solution for $y(x)$ ?
Answer: $-e^{-y}=\frac{1}{3}x^3+x+C$ For $\frac{dy}{dx}=(x^2+1)e^{y}$ , we separate variables to get $e^{-y}dy=(x^2+1)dx$ . Integrating the left side gives $-e^{-y}$ (since $\int e^{-y}dy=-e^{-y}$ ), while the right side gives $\frac{1}{3}x^3+x$ . Therefore, $-e^{-y}=\frac{1}{3}x^3+x+C$ . Choice A incorrectly omits the negative sign when integrating $e^{-y}dy$ , a common error with exponential functions. For separable equations: (1) separate variables, (2) rewrite to standard forms if needed, (3) integrate carefully with correct signs, and (4) verify your antiderivatives.
A quantity $y$ satisfies $\dfrac{dy}{dx}=\dfrac{y}{x^2}$ . What is the general solution for $y(x)$ ?
Answer: $y=Ce^{-1/x}$ For the differential equation dy/dx = y/x², we use separation of variables by rearranging to dy/y = dx/x². Integrating the left side gives ln|y|, while integrating the right side gives -1/x (since ∫x⁻² dx = -x⁻¹). Therefore, ln|y| = -1/x + C₁, and exponentiating both sides yields y = $e^{C₁}$ · $e^{-1/x}$ = $Ce^{-1/x}$ , where C = $e^{C₁}$ is an arbitrary constant. Choice A incorrectly assumes the equation is dy/dx = -y/x instead of y/x². When solving separable equations: (1) separate variables completely, (2) integrate both sides carefully (watch for negative exponents), (3) exponentiate to remove logarithms, and (4) simplify the arbitrary constant.
A chemical amount $y$ changes with $x$ by $\dfrac{dy}{dx}=x(1+y^2)$ . Find the general solution.
Answer: $y=\tan\!\left(\dfrac{x^2}{2}+C\right)$ This differential equation dy/dx = x(1+y²) requires separation of variables to solve. We rearrange to get dy/(1+y²) = x dx, separating y terms on the left and x terms on the right. Integrating the left side gives arctan(y) (since the derivative of arctan(y) is 1/(1+y²)), and integrating the right side gives x²/2. Thus arctan(y) = x²/2 + C, and solving for y yields y = tan(x²/2 + C). Choice A incorrectly stops at the implicit form without solving for y explicitly. When separating variables: (1) isolate dy and y terms on one side, (2) isolate dx and x terms on the other, (3) integrate both sides, and (4) solve for the dependent variable if possible.
A tank cools so that $\frac{dT}{dt}=-0.2(T-70)$ . Which equation represents the general solution for $T(t)$ ?
Answer: $T(t)=70+Ce^{-0.2t}$ Separation of variables is the key skill for solving this Newton's law of cooling differential equation. Rewrite it as dT / (T - 70) = -0.2 dt to separate T and t. Integrate both sides to obtain ln|T - 70| = -0.2 t + K, then exponentiate to get |T - 70| = $e^K$ $e^{-0.2 t}$ . Solving for T yields T(t) = 70 + C $e^{-0.2 t}$ , where C = ± $e^K$ incorporates the constant and sign. Choice B is a tempting distractor because it has a positive exponent, which might result from forgetting the negative sign in the separation step, leading to exponential growth instead of decay. To solve separable equations, always isolate variables, integrate each side, solve for the dependent variable, and incorporate the constant appropriately.
For $y\neq 0$ , suppose $\frac{dy}{dx}=\frac{3x^2}{y^3}$ . What is the general solution?
Answer: $y^4=4x^3+C$ Separation of variables is used to solve this differential equation for y in terms of x. Rearrange as $y^3$ dy = 3 $x^2$ dx. Integrate to obtain (1/4) $y^4$ = $x^3$ + K. Multiply by 4 to get $y^4$ = 4 $x^3$ + C, where C = 4K. Choice B, $y^4$ = 3 $x^3$ + C, is a tempting distractor that might occur from forgetting to multiply the right side by 4 after integrating, thus mismatching the coefficients. To solve separable equations, always isolate variables, integrate each side, solve for the dependent variable, and incorporate the constant appropriately.
A curve satisfies $\dfrac{dy}{dx}=x\sqrt{y}$ . What is the general solution for $y(x)$ ?
Answer: $y=\left(\dfrac{x^2}{4}+C\right)^2$ This problem requires the skill of finding general solutions to differential equations using separation of variables. To solve dy/dx = x sqrt(y), separate variables by writing dy / sqrt(y) = x dx. Integrate both sides: ∫ $y^{-1/2}$ dy = ∫ x dx, yielding 2 sqrt(y) = (1/2) $x^2$ + C. Solve for y: sqrt(y) = (1/4) $x^2$ + K, so y = $((1/4) x^2 + K)^2$ . A tempting distractor is choice C, sqrt(y) = (1/2) $x^2$ + C, which forgets to divide by 2 after integrating the left side. For any separable equation dy/dx = g(x) h(y), rewrite as (1/h(y)) dy = g(x) dx, integrate both sides, solve for y if possible, and include the constant of integration.
A population satisfies $\frac{dP}{dt}=3P(1-P)$ . What is the general solution for $P(t)$ ?
Answer: $\ln\left|\frac{P}{1-P}\right|=3t+C$ This problem requires separation of variables to solve the differential equation $\frac{dP}{dt}=3P(1-P)$ . To separate variables, we rewrite as $\frac{dP}{P(1-P)}=3dt$ . Using partial fractions, $\frac{1}{P(1-P)}=\frac{1}{P}+\frac{1}{1-P}$ , so integrating both sides gives $\ln|P|-\ln|1-P|=3t+C$ . This simplifies to $\ln\left|\frac{P}{1-P}\right|=3t+C$ using logarithm properties. Choice A incorrectly adds the logarithms instead of subtracting them, which would come from an error in the partial fraction decomposition. When separating variables: (1) isolate all terms with the dependent variable on one side, (2) integrate both sides carefully, and (3) use algebraic properties to simplify the result.
A growth model satisfies $\frac{dy}{dx}=\frac{y^2}{1+x}$ . What is the general solution for $y(x)$ ?
Answer: $-\frac{1}{y}=\ln|1+x|+C$ This growth model uses separation of variables on $\frac{dy}{dx}=\frac{y^2}{1+x}$ . Separating gives $\frac{dy}{y^2}=\frac{dx}{1+x}$ . The left side integrates to $-\frac{1}{y}$ (since $\int y^{-2}dy=-y^{-1}+C$ ), while the right side integrates to $\ln|1+x|$ . Therefore, the general solution is $-\frac{1}{y}=\ln|1+x|+C$ . Choice B incorrectly has $\frac{1}{y}$ instead of $-\frac{1}{y}$ , forgetting the negative sign when integrating $y^{-2}$ . For separation success: (1) separate variables completely, (2) remember that $\int x^{-2}dx=-x^{-1}+C$ , and (3) use substitution when needed for expressions like $\frac{1}{1+x}$ .
A moving particle's velocity $v(t)$ satisfies $\frac{dv}{dt}=5t\sqrt{v}$ . What is the general solution for $v(t)$ ?
Answer: $2\sqrt{v}=\frac{5}{2}t^2+C$ This equation $\frac{dv}{dt}=5t\sqrt{v}$ requires separation of variables. We rewrite $\sqrt{v}$ as $v^{1/2}$ and separate to get $v^{-1/2}dv=5t\,dt$ . Integrating the left side using the power rule gives $2v^{1/2}=2\sqrt{v}$ , while the right side gives $\frac{5}{2}t^2$ . Therefore, $2\sqrt{v}=\frac{5}{2}t^2+C$ . Choice B incorrectly omits the coefficient 2 from integrating $v^{-1/2}$ , while choice E incorrectly integrates $5t$ as $5t$ instead of $\frac{5}{2}t^2$ . When solving: (1) convert radicals to powers, (2) separate variables, (3) apply power rule carefully, and (4) simplify the final form.
A chemical concentration satisfies $\frac{dy}{dt}=y^2$ . Which equation gives the general solution for $y(t)$ ?
Answer: $y=-\dfrac{1}{t+C}$ To solve this differential equation, apply the separation of variables technique. Rewrite as dy / $y^2$ = dt. Integrate both sides to get -1/y = t + K. Solving for y gives y = -1 / (t + C), where C = -K or adjusted accordingly. Choice A, y = 1 / (t + C), is a tempting distractor because it omits the negative sign, which arises from the integral of $y^{-2}$ being -1/y, leading to positive reciprocal if ignored. To solve separable equations, always isolate variables, integrate each side, solve for the dependent variable, and incorporate the constant appropriately.
A motion model satisfies $\frac{dv}{dt}=t\sqrt{v}$ . What is the general solution for $v(t)$ ?
Answer: $2\sqrt{v}=\frac{t^2}{2}+C$ This motion model uses separation of variables on $\frac{dv}{dt}=t\sqrt{v}$ . Rewriting as $\frac{dv}{\sqrt{v}}=t\,dt$ and recognizing that $\frac{1}{\sqrt{v}}=v^{-1/2}$ , we integrate both sides. The left side gives $2\sqrt{v}$ (since $\int v^{-1/2}dv=2v^{1/2}+C$ ), while the right side gives $\frac{t^2}{2}$ . Therefore, the general solution is $2\sqrt{v}=\frac{t^2}{2}+C$ . Choice B incorrectly computes the left integral as $\sqrt{v}$ instead of $2\sqrt{v}$ , forgetting to apply the power rule correctly for $n=-1/2$ . For separation success: (1) rewrite radicals as powers, (2) use the power rule carefully, and (3) check your work by differentiating.
A tank's salt amount $S(t)$ satisfies $\frac{dS}{dt}=\frac{3t}{S}$ . What is the general solution for $S(t)$ ?
Answer: $\frac{1}{2}S^2=\frac{3}{2}t^2+C$ This problem uses separation of variables for the equation $\frac{dS}{dt}=\frac{3t}{S}$ . We separate by multiplying both sides by $S$ and $dt$ to get $S\,dS=3t\,dt$ . Integrating both sides yields $\frac{1}{2}S^2=\frac{3}{2}t^2+C$ , where we use the power rule on both sides. Choice A incorrectly omits the coefficient $\frac{1}{2}$ from integrating $S\,dS$ , while choice E incorrectly integrates $3t$ as $3t$ instead of $\frac{3}{2}t^2$ . For separable equations: (1) separate variables completely, (2) integrate each side independently, (3) include all coefficients from integration, and (4) add the constant of integration.
A quantity satisfies $\frac{dy}{dx}=\frac{2y}{x^2}$ for $x\neq 0$ . What is the general solution?
Answer: $y(x)=Ce^{-\frac{2}{x}}$ This differential equation $\frac{dy}{dx}=\frac{2y}{x^2}$ requires separation of variables. We separate by dividing both sides by $y$ to get $\frac{dy}{y}=\frac{2}{x^2}dx=2x^{-2}dx$ . Integrating both sides yields $\ln|y|=2\cdot\frac{x^{-1}}{-1}+C_1=-\frac{2}{x}+C_1$ . Exponentiating gives $|y|=e^{-\frac{2}{x}+C_1}=e^{C_1}e^{-\frac{2}{x}}$ , so $y(x)=Ce^{-\frac{2}{x}}$ where $C=\pm e^{C_1}$ . Choice B, $y(x)=Ce^{\frac{2}{x}}$ , has the wrong sign because it forgets that integrating $x^{-2}$ gives $-x^{-1}$ , not $x^{-1}$ . When separating variables with negative powers: (1) remember that $\int x^n dx = \frac{x^{n+1}}{n+1}$ for $n\neq -1$ , (2) be careful with signs when $n<0$ , (3) exponentiate to solve for the dependent variable, and (4) double-check by differentiating your answer.
A function $y(x)$ satisfies $\frac{dy}{dx}=\frac{2y}{x}$ . What is the general solution for $y(x)$ ?
Answer: $\ln|y|=\ln|x|^2+C$ For $\frac{dy}{dx}=\frac{2y}{x}$ , we use separation of variables to get $\frac{dy}{y}=\frac{2dx}{x}$ . Integrating both sides gives $\ln|y|=2\ln|x|+C$ . Using logarithm properties, $2\ln|x|=\ln|x|^2$ , so we have $\ln|y|=\ln|x|^2+C$ . Choice C shows $y=Cx^2$ , which is the exponential form of our answer but uses $C$ multiplicatively rather than additively in the logarithmic form. For separable equations: (1) separate variables, (2) integrate both sides, (3) use logarithm properties to simplify, and (4) decide whether to leave in logarithmic or exponential form.
A curve satisfies $\dfrac{dy}{dx}=\dfrac{x}{1+y^2}$ . What is the general solution for $y(x)$ ?
Answer: $y+\dfrac{y^3}{3}=\dfrac{x^2}{2}+C$ This problem requires the skill of finding general solutions to differential equations using separation of variables. To solve dy/dx = x / (1 + $y^2$ ), separate variables by writing (1 + $y^2$ ) dy = x dx. Integrate both sides: ∫ (1 + $y^2$ ) dy = ∫ x dx, yielding y + (1/3) $y^3$ = (1/2) $x^2$ + C. This is the implicit general solution. A tempting distractor is choice C, ln|1 + $y^2$ | = (1/2) $x^2$ + C, which incorrectly integrates the left side as a log instead of a polynomial. For any separable equation dy/dx = g(x) h(y), rewrite as (1/h(y)) dy = g(x) dx, integrate both sides, solve for y if possible, and include the constant of integration.
A particle's velocity satisfies $\frac{dv}{dt}=v\cos t$ . What is the general solution for $v(t)$ ?
Answer: $v(t)=Ce^{\sin t}$ The differential equation $\frac{dv}{dt}=v\cos t$ is solved using separation of variables. We divide both sides by $v$ to get $\frac{dv}{v}=\cos t\,dt$ . Integrating both sides yields $\ln|v|=\sin t + C_1$ , since $\int\cos t\,dt=\sin t$ . Exponentiating gives $|v|=e^{\sin t + C_1}=e^{C_1}e^{\sin t}$ , so $v(t)=Ce^{\sin t}$ where $C=\pm e^{C_1}$ . Choice C, $v(t)=Ce^{\cos t}$ , incorrectly assumes that the integral of $\cos t$ is $\cos t$ , when it's actually $\sin t$ . For separation of variables with trigonometric functions: (1) separate the variables, (2) remember that $\int\cos t\,dt=\sin t$ and $\int\sin t\,dt=-\cos t$ , (3) exponentiate to solve for the dependent variable, and (4) express the arbitrary constant clearly.
A quantity satisfies $\frac{dy}{dx}=\frac{y}{x^2}$ . What is the general solution for $y(x)$ ?
Answer: $\ln|y|=-\frac{1}{x}+C$ This problem uses separation of variables on $\frac{dy}{dx}=\frac{y}{x^2}$ . Separating gives $\frac{dy}{y}=\frac{dx}{x^2}$ . The left side integrates to $\ln|y|$ , while the right side integrates to $-\frac{1}{x}$ (since $\int x^{-2}dx=-x^{-1}+C$ ). Therefore, the general solution is $\ln|y|=-\frac{1}{x}+C$ . Choice A incorrectly integrates $\frac{1}{x^2}$ as $\frac{1}{x}$ instead of $-\frac{1}{x}$ , forgetting that $\int x^{-2}dx$ requires the power rule with $n=-2$ . For separation success: (1) separate all instances of each variable, (2) integrate using correct formulas, and (3) remember that $\int x^n dx = \frac{x^{n+1}}{n+1}+C$ for $n\neq -1$ .
A bacteria culture satisfies $\dfrac{dP}{dt}=0.4P$ . What is the general solution for $P(t)$ ?
Answer: $P(t)=Ce^{0.4t}$ This problem requires the skill of finding general solutions to differential equations using separation of variables. To solve dP/dt = 0.4 P, separate variables by writing dP / P = 0.4 dt. Integrate both sides: ∫ (1/P) dP = ∫ 0.4 dt, yielding ln|P| = 0.4 t + C. Exponentiate to get P = $e^{0.4 t + C}$ = C $e^{0.4 t}$ , where C is an arbitrary constant. A tempting distractor is choice A, P(t) = 0.4 t + C, which incorrectly assumes linear growth instead of exponential proportionality. For any separable equation dy/dx = g(x) h(y), rewrite as (1/h(y)) dy = g(x) dx, integrate both sides, solve for y if possible, and include the constant of integration.
A cooling object satisfies $\dfrac{dT}{dt}=-k(T-20)$ . What is the general solution for $T(t)$ ?
Answer: $T(t)=20+Ce^{-kt}$ This problem requires the skill of finding general solutions to differential equations using separation of variables. To solve dT/dt = -k (T - 20), separate variables by writing dT / (T - 20) = -k dt. Integrate both sides: ∫ 1/(T - 20) dT = ∫ -k dt, yielding ln|T - 20| = -k t + C. Exponentiate to get T - 20 = A $e^{-k t}$ , or T = 20 + A $e^{-k t}$ . A tempting distractor is choice B, T(t) = 20 + C $e^{k t}$ , which has a positive exponent and would model heating away from equilibrium instead of approaching it. For any separable equation dy/dx = g(x) h(y), rewrite as (1/h(y)) dy = g(x) dx, integrate both sides, solve for y if possible, and include the constant of integration.
A bacteria culture $B(t)$ satisfies $\frac{dB}{dt}=\frac{B}{1+t^2}$ . What is the general solution for $B(t)$ ?
Answer: $\ln|B|=\arctan(t)+C$ This differential equation $\frac{dB}{dt}=\frac{B}{1+t^2}$ requires separation of variables. We separate to get $\frac{dB}{B}=\frac{dt}{1+t^2}$ . The left side integrates to $\ln|B|$ , while the right side is the standard integral $\arctan(t)$ , giving us $\ln|B|=\arctan(t)+C$ . Choice C incorrectly treats $\frac{1}{1+t^2}$ as the antiderivative rather than recognizing it as the derivative of $\arctan(t)$ . When solving separable equations: (1) separate variables, (2) recognize standard integral forms, (3) integrate both sides, and (4) include the constant of integration.

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