Finding General Solutions: Separation of Variables
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AP Calculus AB › Finding General Solutions: Separation of Variables
A quantity satisfies $\dfrac{dy}{dx}=-\dfrac{3}{x}y$ for $x\ne 0$. What is the general solution?
$y=-3\ln|x|+C$
$\ln|y|=-\dfrac{3}{x}+C$
$y=Ce^{-3x}$
$y=Cx^{-3}$
$y=x^{-3}+C$
Explanation
This problem requires solving a differential equation using separation of variables. Given dy/dx = -(3/x) y for x ≠ 0, separate as dy/y = -(3/x) dx, assuming y ≠ 0. Integrate to ln|y| = -3 ln|x| + C. This simplifies to y = C $x^{-3}$. The choice y = -3 ln|x| + C is incorrect as it stops at the integral without solving for y exponentially. 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 $\dfrac{dy}{dx}=\dfrac{3}{x}y$ for $x\ne 0$. What is the general solution?
$\ln|y|=\dfrac{3}{x}+C$
$y=x^{3}+C$
$y=3\ln|x|+C$
$y=Ce^{3/x}$
$y=Cx^{3}$
Explanation
This problem requires solving a differential equation using separation of variables. Given dy/dx = (3/x) y for x ≠ 0, separate as dy/y = (3/x) dx, assuming y ≠ 0. Integrate: ln|y| = 3 ln|x| + C. Simplify to y = C $x^3$. The choice y = 3 ln|x| + C is incorrect because it omits exponentiation to power form. 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?
$y=Ce^{2/x}$
$\ln|y|=\dfrac{2}{x}+C$
$y=2\ln|x|+C$
$y=Cx^{2}$
$y=x^{2}+C$
Explanation
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 function satisfies $\dfrac{dy}{dx}=\dfrac{y}{x}$. What is the general solution for $x\ne 0$?
$y=x+C$
$y=\ln|x|+C$
$y=Cx$
$\ln|y|=\dfrac{1}{x}+C$
$y=Ce^{x}$
Explanation
This problem requires solving a differential equation using separation of variables. For dy/dx = y/x with x ≠ 0, separate as dy/y = dx/x, assuming y ≠ 0. Integrate: ln|y| = ln|x| + C. Simplify to y = C x. The distractor y = x + C fails by suggesting addition instead of the proportional solution from logs. 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 $\dfrac{dy}{dx}=y\cos x$. What is the general solution for $y(x)$?
$\ln|y|=\cos x+C$
$y=C\sin x$
$y=Ce^{\sin x}$
$y=e^{\sin x}+C$
$y=Ce^{-\sin x}$
Explanation
This problem requires solving a differential equation using separation of variables. With dy/dx = y cos x, separate as dy/y = cos x dx, assuming y ≠ 0. Integrate for ln|y| = sin x + C. Solve to y = C $e^{sin x}$. The option y = $e^{sin x}$ + C is wrong because it places C additively, not multiplicatively. 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 changes by $\dfrac{dy}{dx}=5y$. What is the general solution for $y(x)$?
$\ln|y|=5x$
$y=5e^{x}+C$
$y=5x+C$
$y=e^{5x}+C$
$y=Ce^{5x}$
Explanation
This problem requires solving a differential equation using separation of variables. For dy/dx = 5y, separate as dy/y = 5 dx, assuming y ≠ 0. Integrate: ln|y| = 5x + C. Exponentiate to y = $Ce^{5x}$. The distractor y = $e^{5x}$ + C fails by placing the constant additively, not multiplicatively as required. 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 $\frac{dy}{dx}=-7y$. What is the general solution for $y(x)$?
$y=Ce^{-7x}$
$y=e^{-7x}$
$\ln|y|=-7$
$y=C-7x$
$y=Ce^{7x}$
Explanation
To solve $\frac{dy}{dx}=-7y$ using separation of variables, we divide by $y$ and multiply by $dx$ to get $\frac{dy}{y}=-7dx$. Integrating both sides gives $\ln|y|=-7x+C_1$. Exponentiating yields $|y|=e^{-7x+C_1}=e^{C_1}e^{-7x}$, which becomes $y=Ce^{-7x}$ where $C=\pm e^{C_1}$. Choice B incorrectly uses a positive exponent, missing the negative sign from the differential equation. The key recognition pattern is that $\frac{dy}{dx}=ky$ always yields $y=Ce^{kx}$, with the sign of $k$ preserved in the solution.
A model satisfies $\frac{dy}{dx}=\frac{3x}{1+x^2}y$. What is the general solution for $y(x)$?
$\ln y=\frac{3x}{1+x^2}+C$
$y=\frac{3}{2}\ln(1+x^2)+C$
$y=(1+x^2)^{3/2}$
$y=C(1+x^2)^{3/2}$
$\ln(1+x^2)=\frac{3}{2}y+C$
Explanation
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.
For an investment, $\frac{dA}{dt}=0.08A$. What is the general solution for $A(t)$?
$A(t)=0.08e^{t}$
$A(t)=Ce^{0.08t}$
$A(t)=e^{0.08t}$
$A(t)=C+0.08t$
$\ln A=0.08$
Explanation
The investment equation $\frac{dA}{dt}=0.08A$ is solved through separation of variables. Dividing by $A$ and multiplying by $dt$ gives $\frac{dA}{A}=0.08dt$. Integration yields $\ln|A|=0.08t+C_1$. Exponentiating both sides gives $|A|=e^{0.08t+C_1}=e^{C_1}e^{0.08t}$, resulting in $A=Ce^{0.08t}$ where $C=\pm e^{C_1}$. Choice A incorrectly places the rate 0.08 as a coefficient outside the exponential rather than in the exponent. When solving $\frac{dy}{dt}=ky$, the constant $k$ always appears in the exponent of $e$ in the solution $y=Ce^{kt}$.
For a chemical reaction, $\frac{dy}{dx}=\frac{4}{x}y$. What is the general solution for $y(x)$?
$y=C\ln|x|^{4}$
$\ln|y|=\frac{4}{x}+C$
$y=4\ln|x|+C$
$y=Cx^{4}$
$y=x^{4}$
Explanation
To solve $\frac{dy}{dx}=\frac{4}{x}y$ using separation of variables, we divide both sides by $y$ and multiply by $dx$ to get $\frac{dy}{y}=\frac{4}{x}dx$. Integrating both sides yields $\ln|y|=4\ln|x|+C_1=\ln|x|^4+C_1$. Exponentiating gives $|y|=e^{\ln|x|^4+C_1}=e^{C_1}|x|^4$, which simplifies to $y=Cx^4$ where $C=\pm e^{C_1}$. Choice A incorrectly integrates the right side as if $y$ were not present in the original equation. When the coefficient of $y$ contains only functions of the independent variable, separation of variables transforms the exponential into a power function.