AP Calculus AB Unit 7: Differential Equations — Free Review Games.

A slope field visualizes a differential equation by drawing small line segments at each point with the slope given by \(\frac{dy}{dx}\) at that point.

Integrate: \(y = x^3 + C\). Using \(y(0) = 5\): \(5 = 0 + C\), so \(C = 5\). Solution: \(y = x^3 + 5\).

\(\frac{dy}{dx} = ky\) means the rate of change is proportional to the current value. Its solution is \(y = Ce^{kx}\), exponential growth when \(k > 0\).

Separation of variables rewrites \(\frac{dy}{dx} = f(x)g(y)\) as \(\frac{dy}{g(y)} = f(x)dx\), allowing each side to be integrated independently.

Integrating both sides: \(y = x^2 + C\).

Separate: \(\frac{dy}{y} = dx\). Integrate: \(\ln|y| = x + C\). \(y = Ae^x\). \(y(0) = 3\): \(A = 3\). Solution: \(y = 3e^x\).

Solution: P = 1000e^(0.05t). P(10) = 1000e^(0.5) = approximately 1649.

Separate: \(y \, dy = x \, dx\). Integrate: \(\frac{y^2}{2} = \frac{x^2}{2} + C_1\). Simplifying: \(y^2 = x^2 + C\).

Horizontal segments mean the slope is zero. These occur where \(\frac{dy}{dx} = 0\), indicating potential maxima, minima, or equilibrium solutions.

The logistic equation models population growth that slows as P approaches carrying capacity M. Growth rate decreases as resources become limited.

Separate: \(\frac{dy}{y} = 2x \, dx\). Integrate: \(\ln|y| = x^2 + C\). \(y = Ae^{x^2}\). \(y(0) = 1\): \(A = 1\). Solution: \(y = e^{x^2}\).

$y_{new} = y_{old} + h \cdot f(x_{old}, y_{old}) = 1 + 0.1 \cdot (0 + 1) = 1 + 0.1 = 1.1$.

Half-life: 50 = 100e^(-5k), k = ln(2)/5 = 0.1386. So A = 100e^(-0.1386t).

The logistic growth rate is maximized at P = M/2. Here M = 100, so maximum growth occurs at P = 50.

Separate: \(\frac{dy}{1+y^2} = dx\). Integrate: \(\arctan(y) = x + C\). \(y(0) = 0\): \(C = 0\). So \(y = \tan(x)\).

A slope field draws a short line segment at each \((x,y)\) whose slope equals \(f(x,y)\), directly visualizing the derivative defined by the differential equation. The distractor 'The value of \(y\) at that point' is wrong because slope fields encode rate of change, not the function's height, which is why segments alone cannot tell you the exact \(y\)-value. On the exam, remember that slope fields let you sketch approximate solution curves by following the direction indicated at each point.

A negative proportionality constant in \(\frac{dy}{dx}=ky\) means the rate of change is always opposite in sign to \(y\), causing \(y\) to shrink toward zero, which is the hallmark of exponential decay. The distractor \(\frac{dy}{dx}=2y\) is wrong because a positive constant makes \(y\) grow, producing exponential growth instead. Recognizing the sign of \(k\) in \(y'=ky\) instantly tells you whether the model is growth or decay.

Separating variables in \(\frac{dy}{dx}=ky\) gives \(\frac{dy}{y}=k\,dx\), and integrating both sides produces \(\ln|y|=kx+C_1\), which exponentiates to \(y=Ce^{kx}\). The distractor \(y=kx+C\) is wrong because that is the solution to \(\frac{dy}{dx}=k\), a constant rate, not a rate proportional to \(y\). This exponential solution form is the foundation for every growth and decay model on the AP exam.

A particular solution is obtained by plugging a given initial condition, such as \(y(x_0)=y_0\), into the general solution to solve for the specific value of the constant \(C\). The distractor 'A solution valid only near x=0' is wrong because particular solutions are valid over whatever domain the differential equation and initial condition support, not restricted to \(x=0\). Always use the initial condition immediately after finding the general solution to pin down the exact function.

Integrating \(\frac{dy}{dx}=6\) gives \(y=6x+C\), and substituting \(x=1,y=2\) yields \(2=6+C\), so \(C=-4\), giving \(y=6x-4\). The distractor \(y=6x+2\) is wrong because it incorrectly reuses the given \(y\)-value as the constant instead of solving for \(C\) algebraically. Always substitute the initial condition into the general antiderivative to correctly solve for \(C\).

A horizontal segment has zero slope, and since the slope field encodes \(\frac{dy}{dx}\), a horizontal segment simply means the derivative equals zero at that location. The distractor '\(y=0\) at those points' is wrong because horizontal slope depends on the differential equation's formula, which need not force \(y\) to be zero. Horizontal segments mark locations where the solution curve momentarily flattens, not necessarily where the curve crosses the x-axis.

Before integrating, you must algebraically isolate all \(y\) terms with \(dy\) on one side and all \(x\) terms with \(dx\) on the other, which is the defining step of separation of variables. The distractor 'Integrate both sides immediately' is wrong because integrating before separating would mix variables and make the antiderivatives impossible to evaluate correctly. Mastering this separation step first is essential before any integration can proceed.

Since the solution is \(y=Ce^{-3x}\) with \(C>0\), the negative exponent causes the function to shrink continuously as \(x\) increases, approaching but never reaching zero. The distractor 'y grows without bound' is wrong because that behavior occurs only when the exponential rate constant is positive, not negative. Any equation of the form \(y'=ky\) with \(k<0\) always produces decay toward zero regardless of the positive starting value.

Separating variables in \(\frac{dP}{dt}=kP\) and integrating gives \(\ln P = kt + C\), which exponentiates to \(P(t)=P_0e^{kt}\) where \(P_0\) is the initial population. The distractor \(P(t)=kt+P_0\) is wrong because that is a linear model, which would only solve \(\frac{dP}{dt}=k\) (a constant rate), not a rate proportional to \(P\) itself. This exponential form is the standard model for unrestricted population or investment growth on the AP exam.

An initial condition gives one known point on the solution curve, which is substituted into the general solution to algebraically solve for the arbitrary constant \(C\). The distractor 'The slope of the tangent line at \(x=0\) only' is wrong because the initial condition specifies a \(y\)-value at a given \(x\), not a derivative value, unless the problem states otherwise. Every particular solution problem requires exactly this substitution step to move from the general family of solutions to one specific curve.

Integrating \(5x^4\) gives \(x^5+C\), and applying \(y(0)=1\) gives \(0+C=1\), so \(C=1\), resulting in \(y=x^5+1\). The distractor \(y=5x^5+1\) is wrong because it fails to divide the coefficient by the increased exponent correctly, essentially skipping the power rule for antiderivatives. Always apply the reverse power rule, \(\int x^n dx = \frac{x^{n+1}}{n+1}+C\), carefully before applying the initial condition.

A differential equation is an equation that involves a derivative, such as \(\frac{dy}{dx}\), relating a function to its rate of change. The distractor \(y=x^2+3x\) is wrong because it is simply an explicit function with no derivative present, making it an ordinary algebraic equation rather than a differential one. Recognizing the presence of \(\frac{dy}{dx}\) or \(y'\) is the quickest way to identify a differential equation on sight.

Plugging \((3,3)\) into \(\frac{dy}{dx}=x-y\) gives \(3-3=0\), so the slope at that point is zero, meaning the segment is horizontal. The distractor 'vertical' is wrong because a vertical segment would require an undefined (infinite) slope, which never occurs for this polynomial expression. Always substitute the coordinates directly into the differential equation to determine the exact segment orientation at any given point.

The solution to \(\frac{dy}{dx}=-0.3y\) with \(y(0)=50\) is \(y=50e^{-0.3x}\), so evaluating at \(x=4\) gives \(y=50e^{-1.2}\approx15.1\). The distractor \(50e^{-0.3}\approx37.0\) is wrong because it mistakenly plugs in \(x=1\) instead of multiplying the rate constant by the full time value \(x=4\). Always multiply the decay constant by the entire elapsed time before exponentiating to evaluate the exponential model at a later time.

Separating gives \(\frac{dy}{y}=x^2\,dx\), and integrating both sides yields \(\ln|y|=\frac{x^3}{3}+C_1\), which exponentiates to \(y=Ce^{x^3/3}\). The distractor \(y=Ce^{x^2/2}\) is wrong because it integrates \(x^2\) as if it were \(x\), incorrectly applying the power rule to the exponent. Careful application of \(\int x^n dx=\frac{x^{n+1}}{n+1}\) is essential before exponentiating to recover \(y\).

Since \(T-T_s\) is positive and \(k\) is negative, \(\frac{dT}{dt}\) is negative, meaning the temperature steadily decreases, and as \(T\) approaches \(T_s\) the rate of change approaches zero, so the object cools toward the ambient temperature. The distractor 'decrease without bound past \(T_s\)' is wrong because the driving term \((T-T_s)\) shrinks to zero as \(T\) nears \(T_s\), preventing the temperature from dropping below the surrounding value. This asymptotic approach toward equilibrium is the key qualitative behavior of Newton's Law of Cooling.

Separating variables gives \(\frac{dy}{y}=\frac{dx}{x^2}\), and integrating both sides yields \(\ln|y|=-\frac{1}{x}+C_1\), which exponentiates to \(y=Ce^{-1/x}\). The distractor \(y=Ce^{1/x}\) is wrong because it drops the negative sign that results from integrating \(x^{-2}\), which produces \(-x^{-1}\), not \(x^{-1}\). Careful sign-tracking during integration of negative powers of \(x\) is critical for arriving at the correct exponential form.

In the model \(\frac{dP}{dt}=kP\), the constant \(k\) directly represents the continuous relative growth rate, so \(k=0.08\) corresponds to an approximate 8% annual growth rate. The distractor '0.08%' is wrong because it misplaces the decimal, confusing the decimal form of the rate with a percentage that is a hundred times smaller. Always convert the constant \(k\) to a percentage by multiplying by 100 to interpret growth or decay rates correctly.

Setting \(A_0e^{-0.02t}=0.5A_0\) and solving gives \(t=\frac{\ln2}{0.02}\approx34.7\) years, which is the half-life formula derived directly from the exponential decay solution. The distractor '\(t=50\) years' is wrong because it only inverts the decay constant \(\frac{1}{0.02}\) without accounting for the necessary factor of \(\ln 2\) from solving the exponential equation. Remember that half-life always equals \(\frac{\ln 2}{k}\) for a decay constant \(k\), not simply \(\frac{1}{k}\).

Separating gives \(\frac{dy}{y}=x\,dx\), integrating yields \(\ln y=\frac{x^2}{2}+C\), and applying \(y(0)=2\) gives \(y=2e^{x^2/2}\), so \(y(1)=2e^{0.5}\approx3.30\). The distractor \(2e^{1}\approx5.44\) is wrong because it forgets to divide the exponent from integrating \(x\), which produces \(\frac{x^2}{2}\) rather than \(x^2\). Careful integration of the \(x\)-side is just as important as the \(y\)-side when separating variables.

The solution is \(A(t)=1000e^{0.06t}\), so \(A(10)=1000e^{0.6}\approx\\)1822$, reflecting continuous compounding at rate 6%. The distractor '\(\\)1060$' is wrong because it applies the rate only linearly for one year rather than compounding continuously over the full ten-year period. Continuous compound growth problems always require exponentiating the product of rate and time, not simply multiplying them.

Differentiating \(y=3e^{2x}\) gives \(\frac{dy}{dx}=6e^{2x}\), which equals \(2(3e^{2x})=2y\), confirming the function satisfies the differential equation. The distractor 'True, because \(y(0)=3\)' is wrong because matching a single initial value does not verify that the function satisfies the derivative relationship for all \(x\). To verify any proposed solution, always differentiate it and substitute back into the original differential equation to check the identity holds.

In \(\frac{dy}{dx}=-y\), the slope equals zero exactly when \(y=0\), and as \(y\) grows larger the slope becomes increasingly negative, matching the described pattern of steepening negative segments. The distractor \(\frac{dy}{dx}=x\) is wrong because its slope depends only on \(x\), not \(y\), so it would produce identical slopes along any vertical line regardless of \(y\)-value. Reading how slope magnitude changes with \(y\) versus \(x\) is a key skill for matching a slope field to its equation.

Separating gives \(y\,dy=\cos x\,dx\), integrating yields \(\frac{y^2}{2}=\sin x+C\), and using \(y(0)=1\) gives \(C=\frac{1}{2}\), so \(y^2=2\sin x+1\), and \(y=\sqrt{2\sin x+1}\). The distractor \(y=\sin x+1\) is wrong because it fails to square the \(y\) term properly during separation, effectively skipping the necessary square root step. Whenever \(y\) appears to the first power on the separated side, remember to solve for \(y\) using a square root after integrating.

Rewriting as \(e^y\,dy=e^x\,dx\) and integrating gives \(e^y=e^x+C\), and solving for \(y\) gives \(y=\ln(e^x+C)\). The distractor \(y=x+C\) is wrong because it ignores the exponential relationship entirely and treats the equation as if it were linear in \(x\) alone. Rewriting \(e^{x-y}\) as \(e^x\cdot e^{-y}\) is the key algebraic move that makes this equation separable.

The solution is \(Q(t)=20e^{-0.1t}\), so \(Q(5)=20e^{-0.5}\approx12.1\) kg, reflecting exponential decay of the salt amount once fresh water is no longer entering. The distractor '\(\approx10.0\) kg' is wrong because it assumes salt decreases linearly by half over 5 time units rather than following the exponential decay curve. Mixing problems that reduce to \(Q'=-kQ\) always follow the same exponential decay solution as radioactive decay.

Using \(y=10e^{kx}\) and substituting \(y(3)=40\) gives \(4=e^{3k}\), so \(k=\frac{\ln4}{3}\approx0.462\). The distractor \(k=\frac{4}{3}\) is wrong because it treats the ratio of \(y\)-values as the value of \(k\) directly instead of taking the natural log of that ratio and dividing by the elapsed time. Always isolate the exponential term and apply a natural logarithm before solving for the rate constant \(k\).

The solution is \(y=2e^{3x}\), so evaluating at \(x=1\) gives \(y(1)=2e^3\approx40.2\). The distractor \(2e^{1}\approx5.44\) is wrong because it omits the coefficient 3 from the exponent, treating the equation as if \(k=1\) instead of \(k=3\). Always carry the full rate constant into the exponent of \(e^{kx}\) before evaluating at a specific \(x\)-value.

Setting \(2P_0=P_0e^{3k}\) gives \(2=e^{3k}\), so \(k=\frac{\ln2}{3}\approx0.231\) per hour. The distractor \(\frac{3}{\ln2}\approx4.33\) is wrong because it inverts the correct ratio, dividing time by the natural log instead of dividing the natural log by time. Doubling-time problems always require solving \(e^{kt}=2\) for \(k\) given the specified doubling period \(t\).

The equation \(\frac{dy}{dx}=xy\) can be rewritten as a product of a function of \(x\) alone and a function of \(y\) alone, allowing the variables to be separated into \(\frac{dy}{y}=x\,dx\). The distractor \(\frac{dy}{dx}=x+y\) is wrong because addition of \(x\) and \(y\) cannot be factored into a pure product of separate functions of each variable. A differential equation is separable only when the right side can be written as \(g(x)h(y)\), a product rather than a sum.

Since the right side of the differential equation contains only \(x\), the slope at any point depends solely on the \(x\)-coordinate, so all points sharing the same \(x\)-value have identical segment slopes regardless of \(y\). The distractor 'are the same at every point' is wrong because the slope clearly changes as \(x\) changes, only remaining constant along vertical lines of fixed \(x\). When a differential equation's right side involves only one variable, the slope field's pattern is uniform along lines of that variable held constant.

Rewriting as \(\frac{dy}{5-y}=dx\) and solving gives \(y=5-5e^{-x}\), and as \(x\to\infty\) the exponential term vanishes, leaving \(y\to5\). The distractor \(y=0\) is wrong because that is only the starting value at \(x=0\), not the long-term behavior as the exponential decay term disappears. Equations of the form \(y'=k(L-y)\) always drive solutions toward the equilibrium value \(L\) as \(x\) increases.

Separating gives \(\frac{dy}{y^2}=x\,dx\), integrating yields \(-\frac{1}{y}=\frac{x^2}{2}+C\), and applying \(y(1)=1\) gives \(C=-1.5\), so at \(x=2\), \(-\frac{1}{y}=0.5\), giving \(y=-2\). The distractor \(y=2\) is wrong because it ignores the sign that results from solving \(-\frac{1}{y}=0.5\), which forces \(y\) to be negative rather than positive. Multi-step separable problems require careful algebraic solving for \(y\) at the very end, including sign checks.

The first Euler step gives \(y_1=1+0.2(0+1)=1.2\) at \(x=0.2\), and the second step gives \(y_2=1.2+0.2(0.2+1.2)=1.48\) at \(x=0.4\). The distractor '\(1.28\)' is wrong because it likely uses only one Euler step or miscalculates the slope at the intermediate point \((0.2,1.2)\). Euler's method requires recomputing the slope at each new approximated point before taking the next step forward.

The logistic growth rate is maximized exactly at half the carrying capacity, \(P=\frac{M}{2}=\frac{200}{2}=100\), where the product \(P(1-P/M)\) reaches its largest value. The distractor \(P=200\) is wrong because that is the carrying capacity itself, where the growth rate actually equals zero since the population stops growing. Always remember that the inflection point and maximum growth rate of a logistic curve occur at half the carrying capacity, not at the capacity itself.

Separating gives \(y\,dy=(x+1)\,dx\), integrating yields \(\frac{y^2}{2}=\frac{x^2}{2}+x+C\), and applying \(y(0)=-2\) gives \(y^2=x^2+2x+4\), whose discriminant \(4-16<0\) shows the expression under the square root stays positive for every real \(x\). The distractor '\(x>-1\) only' is wrong because it assumes the quadratic could become negative near \(x=-1\), but completing the square shows \(x^2+2x+4=(x+1)^2+3\), which never reaches zero. Checking the discriminant or completing the square is essential to correctly determine domain restrictions for implicit solutions.

With \(k=\frac{\ln2}{8}\approx0.0866\), solving \(e^{-kt}=0.1\) gives \(t=\frac{\ln10}{k}\approx26.6\) years, showing significantly more time is needed to reach 10% than a single half-life. The distractor '\(\approx16.0\) years' is wrong because it merely doubles the half-life, incorrectly assuming two half-lives leave 10% remaining when in fact two half-lives leave 25% remaining. Decay-to-a-given-fraction problems always require solving \(e^{-kt}=\text{fraction}\) directly rather than counting whole half-life intervals.

Solving \(0.8e^{-0.05t}=0.2\) gives \(e^{-0.05t}=0.25\), so \(t=\frac{\ln4}{0.05}\approx27.7\) minutes, following the same exponential decay structure as radioactive decay problems. The distractor '\(\approx13.9\) minutes' is wrong because it corresponds to halving the concentration once, not the required reduction to one-quarter of the initial value. Always solve the full exponential equation for the exact target ratio rather than approximating with a single half-life step.

Below \(P=M/2\) the growth rate is increasing, giving concave up behavior, while above \(P=M/2\) the growth rate begins decreasing as the population approaches the carrying capacity, giving concave down behavior, which is exactly why \(P=M/2\) is the inflection point. The distractor 'concave down before, concave up after' is wrong because it reverses the actual shape of the S-curve, which grows fastest and steepest just before leveling off, not the opposite. The logistic curve's characteristic S-shape always transitions from concave up to concave down precisely at half the carrying capacity.

With \(k=\frac{\ln3}{5}\), solving \(e^{kt}=10\) gives \(t=\frac{\ln10}{k}=\frac{5\ln10}{\ln3}\approx10.5\) years, since the growth constant found from the tripling condition must be reused to solve for the new target ratio. The distractor '\(\approx15.0\) years' is wrong because it assumes reaching 10 times takes three tripling periods, but \(3^3=27\), which overshoots the target of 10 rather than matching it exactly. Multi-step growth problems require deriving \(k\) from the given condition first, then reusing that exact value to solve for the new time.

Separating gives \(\frac{dy}{y}=\cos x\,dx\), integrating yields \(\ln y=\sin x+C\), and applying \(y(0)=2\) gives \(y=2e^{\sin x}\), so \(y(\pi/2)=2e^{\sin(\pi/2)}=2e^1\approx5.44\). The distractor \(2e^{-1}\approx0.74\) is wrong because it incorrectly uses a negative exponent, perhaps confusing \(\sin(\pi/2)=1\) with \(-1\). Evaluating trigonometric separable solutions requires correctly computing the trig function value at the specified point before exponentiating.

The solution is \(T=20+70e^{-kt}\), and substituting \(T(10)=60\) gives \(40=70e^{-10k}\), so \(k=\frac{\ln(7/4)}{10}\approx0.056\) per minute. The distractor '\(k\approx0.143\)' is wrong because it likely inverts the ratio inside the logarithm, using \(\ln(4/7)\) without the correct sign adjustment or dividing incorrectly. Newton's Law of Cooling problems require solving for \(k\) using the shifted variable \(T-T_s\), not the raw temperature values.

Setting up the ratio \(\frac{A_0e^{-0.1t}}{A_0e^{-0.2t}}=e^{0.1t}=2\) and solving gives \(t=\frac{\ln2}{0.1}\approx6.93\) years, since the ratio of two exponentials with different decay rates simplifies to a single exponential in the difference of the rates. The distractor '\(\approx3.47\) years' is wrong because it divides \(\ln2\) by the sum or difference incorrectly, effectively halving the correct time value. Comparing two exponential decay processes always reduces to solving \(e^{(k_B-k_A)t}=\text{ratio}\) for the elapsed time.

Substituting \(y=x\) into \(\frac{dy}{dx}=y-x\) gives a slope of zero, so every solution curve has a horizontal tangent line exactly at the moment it crosses this line, though the curve continues on with nonzero slope elsewhere. The distractor 'All solution curves converge to \(y=x\)' is wrong because the curves only momentarily flatten while crossing \(y=x\), they do not settle onto or follow that line afterward. Setting the right-hand side of a differential equation equal to zero identifies where solution curves have horizontal tangents, which is a powerful tool for sketching slope field behavior.

Separating gives \((1+y^2)\,dy=2x\,dx\), and integrating both sides yields \(y+\frac{y^3}{3}=x^2+C\), and applying \(y(0)=0\) gives \(C=0\), so the implicit solution is \(y+\frac{y^3}{3}=x^2\). The distractor \(\arctan y = x^2\) is wrong because that antiderivative would apply only if the original equation had \(y'=\frac{2x}{1+y^2}\) integrated with respect to \(y\) on the left as \(\frac{1}{1+y^2}\), not \((1+y^2)\) multiplied out as it is here. Distinguishing between \(\frac{1}{1+y^2}\), which integrates to \(\arctan y\), and \((1+y^2)\), which integrates term by term, is a common and important distinction in separable differential equations.