Initial Conditions and Separation of Variables - AP Calculus BC

A function satisfies $\dfrac{dy}{dx}=\dfrac{y}{(x+2)}$ with $y(0)=6$ ; which is the particular solution?
Answer: $y=3(x+2)$ This problem requires applying the initial condition to find the particular solution after solving the differential equation using separation of variables. First, separate variables: dy/y = dx/(x + 2), then integrate both sides to get ln|y| = ln|x + 2| + C. Exponentiate to solve for y: y = $e^C$ (x + 2), or y = K (x + 2). Use the initial condition y(0) = 6: 6 = K * 2, so K = 3, yielding y = 3 (x + 2). A tempting distractor is y = 6 (x + 2), which might result from misapplying the initial condition. Always plug the initial condition into the general solution to determine the constant C and verify by differentiating back.
A chemical concentration satisfies $\frac{dy}{dx}=xy$ with $y(0)=5$ . Which function equals $y(x)$ ?
Answer: $y(x)=5e^{x^2/2}$ The skill demonstrated in this problem is applying an initial condition to find the particular solution of a separable differential equation. The general solution to dy/dx = x y is y(x) = C $e^{x^2/2}$ , found by separating variables and integrating both sides. Substituting the initial condition y(0) = 5 yields 5 = C $e^{0}$ , so C = 5. Thus, the particular solution is y(x) = 5 $e^{x^2/2}$ . A tempting distractor is y(x) = 5 $e^{-x^2/2}$ , which uses a negative sign in the exponent, but it fails because the integration of x dx gives a positive $x^2$ /2. Always substitute the initial values into the general solution to solve for the arbitrary constant and identify the unique function satisfying both the differential equation and the initial condition.
A curve satisfies $\frac{dy}{dx}=(1+y^{2})x$ with $y(0)=0$ . Which particular solution is correct?
Answer: $y=\tan\left(\frac{x^{2}}{2}\right)$ This problem requires applying an initial condition after solving a separable differential equation. Separating variables in $\frac{dy}{dx}=(1+y^2)x$ gives $\frac{dy}{1+y^2}=x\,dx$ . The left side integrates to $\arctan(y)$ and the right to $\frac{x^2}{2}+C$ . So $\arctan(y)=\frac{x^2}{2}+C$ , giving $y=\tan\left(\frac{x^2}{2}+C\right)$ . Using $y(0)=0$ : $0=\tan\left(\frac{0^2}{2}+C\right)=\tan(C)$ , so $C=0$ . Therefore, $y=\tan\left(\frac{x^2}{2}\right)$ . Choice E shows the general solution without applying the initial condition to find $C=0$ . When working with inverse trigonometric functions, remember that $\tan(0)=0$ , making $C=0$ in this case.
Given $\frac{dy}{dx}=\frac{x}{y}$ and $y(2)=3$ , which equation defines the particular solution?
Answer: $y^2=x^2+5$ This problem involves applying an initial condition to an implicit solution of a separable differential equation. Separating variables in dy/dx = x/y gives y dy = x dx, which integrates to y²/2 = x²/2 + C, or y² = x² + 2C. Letting K = 2C, we have y² = x² + K. To find K, substitute the initial condition y(2) = 3: 3² = 2² + K, so 9 = 4 + K, giving K = 5. The particular solution is y² = x² + 5. Choice B incorrectly subtracts 5, likely from a sign error when solving for the constant. Always verify your constant by substituting the initial condition back into your solution.
A solution $y$ satisfies $\frac{dy}{dx}=xy$ and $y(0)=3$ . What is the particular solution?
Answer: $y=3e^{x^{2}/2}$ This problem requires applying an initial condition after solving a separable differential equation. Separating variables in $\frac{dy}{dx}=xy$ gives $\frac{dy}{y}=x\,dx$ . Integrating both sides yields $\ln|y|=\frac{x^2}{2}+C$ , so $y=Ae^{x^2/2}$ where $A=e^C$ . Using the initial condition $y(0)=3$ : $3=Ae^{0^2/2}=Ae^0=A$ , so $A=3$ . Therefore, the particular solution is $y=3e^{x^2/2}$ . Choice C incorrectly integrates to get $e^{x^2}$ instead of $e^{x^2/2}$ , forgetting the factor of $\frac{1}{2}$ from integration. When solving separable equations, carefully integrate both sides and then apply the initial condition to find the constant.
A function satisfies $\frac{dy}{dx}=\frac{4}{y}$ with $y(1)=3$ and $y>0$ . Find $y(x)$ .
Answer: $y=\sqrt{8x+1}$ This problem involves applying an initial condition to find a particular solution. Separating variables in $\frac{dy}{dx}=\frac{4}{y}$ gives $y\,dy=4\,dx$ . Integrating: $\frac{y^2}{2}=4x+C$ , so $y^2=8x+2C$ . Let $K=2C$ , then $y=\sqrt{8x+K}$ (taking positive root since $y>0$ ). Using the initial condition $y(1)=3$ : $3=\sqrt{8(1)+K}=\sqrt{8+K}$ . Squaring: $9=8+K$ , so $K=1$ . Therefore, $y=\sqrt{8x+1}$ . Choice B incorrectly calculates $K=9$ by forgetting that $8(1)=8$ must be included before adding $K$ . When solving implicit equations with initial conditions, substitute both $x$ and $y$ values to find the constant.
A function satisfies $\dfrac{dy}{dx}=\dfrac{x^2}{y}$ with $y(0)=4$ ; which is the particular solution?
Answer: $y=\sqrt{\dfrac{2}{3}x^3+16}$ This problem requires applying the initial condition to find the particular solution after solving the differential equation using separation of variables. First, separate variables: y dy = $x^2$ dx, then integrate both sides to get (1/2) $y^2$ = (1/3) $x^3$ + C. Solve for y: $y^2$ = (2/3) $x^3$ + 2C, or y = sqrt((2/3) $x^3$ + D). Use the initial condition y(0) = 4: 16 = D, so D = 16, yielding y = sqrt((2/3) $x^3$ + 16). A tempting distractor is y = sqrt((1/3) $x^3$ + 16), which might result from incorrect coefficient in integration. Always plug the initial condition into the general solution to determine the constant C and verify by differentiating back.
If $\frac{dy}{dx}=xy$ and $y(0)=3$ , what is the particular solution $y(x)$ ?
Answer: $y(x)=3e^{x^2/2}$ This problem involves using an initial condition to determine the particular solution of a separable differential equation. Separating variables in dy/dx = xy gives dy/y = x dx, which integrates to ln|y| = x²/2 + C₁. Exponentiating both sides yields y = $Ce^{x²/2}$ , where C = ± $e^{C₁}$ . Applying the initial condition y(0) = 3: 3 = $Ce^{0²/2}$ = $Ce^0$ = C·1 = C. Thus C = 3 and the particular solution is y(x) = $3e^{x²/2}$ . Choice C omits the constant multiplier, failing to satisfy the initial condition since it would give y(0) = 1 instead of 3. Remember that initial conditions determine the specific constant in the general solution family.
A function satisfies $\dfrac{dy}{dx}=\dfrac{x}{y^3}$ with $y(0)=2$ ; which $y(x)$ is correct?
Answer: $y=\sqrt[4]{2x^2+16}$ This problem requires applying the initial condition to find the particular solution after solving the differential equation using separation of variables. First, separate variables: $y^3$ dy = x dx, then integrate both sides to get (1/4) $y^4$ = (1/2) $x^2$ + C. Solve for y: $y^4$ = 2 $x^2$ + 4C, or y = $(2 x^2 + D)^{1/4}$ . Use the initial condition y(0) = 2: 2 = $(0 + D)^{1/4}$ , so D = 16, yielding y = $(2 x^2 + 16)^{1/4}$ . A tempting distractor is y = $(2 x^2 + 8)^{1/3}$ , which might result from incorrect exponents in integration. Always plug the initial condition into the general solution to determine the constant C and verify by differentiating back.
A cooling object satisfies $\frac{dT}{dt}=-\frac{1}{10}(T-20)$ with $T(0)=80$ . Which $T(t)$ is correct?
Answer: $T(t)=20+60e^{-t/10}$ This problem involves Newton's law of cooling with an initial condition. The differential equation dT/dt = -(1/10)(T-20) has general solution T(t) = 20 + $Ce^{-t/10}$ , where 20 is the ambient temperature. Applying the initial condition T(0) = 80: 80 = 20 + $Ce^0$ = 20 + C, so C = 60. Therefore, the particular solution is T(t) = 20 + $60e^{-t/10}$ . Choice A (T(t) = 20 + $60e^{-t}$ ) has an incorrect decay rate, missing the factor of 1/10 in the exponent. For cooling problems, the solution approaches the ambient temperature exponentially, and initial conditions determine the temperature difference at t = 0.
A function satisfies $\dfrac{dy}{dx}=\dfrac{y}{x\ln x}$ for $x>1$ with $y(e)=2$ ; find $y(x)$ .
Answer: $y=2\ln x$ This problem requires applying the initial condition to find the particular solution after solving the differential equation using separation of variables. First, separate variables: dy/y = dx/(x ln x), then integrate both sides to get ln|y| = ln|ln x| + C. Exponentiate to solve for y: y = $e^C$ ln x, or y = K ln x. Use the initial condition y(e) = 2: 2 = K ln e = K * 1, so K = 2, yielding y = 2 ln x. A tempting distractor is y = 2 ln(ln x), which might result from incorrectly integrating the right side twice. Always plug the initial condition into the general solution to determine the constant C and verify by differentiating back.
A function satisfies $\frac{dy}{dx}=\frac{x}{1+y^2}$ with $y(0)=1$ . Which is the particular solution?
Answer: $y+\frac{y^3}{3}=\frac{x^2}{2}+\frac{4}{3}$ This problem requires applying an initial condition to an implicit solution. Separating variables in dy/dx = x/(1+y²) gives (1+y²)dy = x dx. Integrating yields y + y³/3 = x²/2 + C. Applying the initial condition y(0) = 1: 1 + 1³/3 = 0²/2 + C gives 1 + 1/3 = 0 + C, so C = 4/3. Therefore, the particular solution is y + y³/3 = x²/2 + 4/3. Choice C (y + y³/3 = x²/2 + 1/3) incorrectly calculates C = 1/3, likely from arithmetic error when evaluating 1 + 1/3. When working with implicit solutions, carefully evaluate all terms at the initial condition to find the correct constant.
A particle's speed satisfies $\frac{dv}{dt}=-v^3$ with $v(0)=2$ . Which function gives $v(t)$ for $t\ge0$ ?
Answer: $v(t)=\frac{2}{\sqrt{1+8t}}$ The skill demonstrated in this problem is applying an initial condition to find the particular solution of a separable differential equation. The general solution to dv/dt = - $v^3$ is v(t) = 1 / sqrt(2 t + C), derived by separating variables, integrating with power rules, and solving for v (taking the positive root since v(0) > 0). Substituting v(0) = 2 gives 2 = 1 / sqrt(C), so C = 1/4. Thus, the particular solution is v(t) = 2 / sqrt(1 + 8 t) after simplifying. A tempting distractor is v(t) = 2 / sqrt(1 - 8 t), which uses a negative sign, but it fails because the differential equation's negative rate leads to a positive term inside the square root for decreasing v. Always substitute the initial values into the general solution to solve for the arbitrary constant and identify the unique function satisfying both the differential equation and the initial condition.
A function satisfies $\dfrac{dy}{dx}=y^2$ with $y(0)=-2$ ; find the particular solution $y(x)$ .
Answer: $y=\dfrac{-2}{1+2x}$ This problem requires applying the initial condition to find the particular solution after solving the differential equation using separation of variables. First, separate variables: dy/ $y^2$ = dx, then integrate both sides to get -1/y = x + C. Solve for y: y = -1/(x + C). Use the initial condition y(0) = -2: -2 = -1/C, so C = 1/2, yielding y = -1/(x + 1/2) = -2/(2x + 1). A tempting distractor is y = -2/(1 - 2x), which might result from sign error in constant. Always plug the initial condition into the general solution to determine the constant C and verify by differentiating back.
A culture grows with $\frac{dP}{dt}=0.2P$ and $P(0)=50$ . Which function gives $P(t)$ ?
Answer: $P(t)=50e^{0.2t}$ This problem requires applying an initial condition to find the particular solution of a differential equation. The general solution to dP/dt = 0.2P is P(t) = $Ce^{0.2t}$ , where C is an arbitrary constant. To find C, we substitute the initial condition P(0) = 50: 50 = $Ce^{0.2·0}$ = $Ce^0$ = C·1 = C. Therefore, C = 50 and the particular solution is P(t) = $50e^{0.2t}$ . Choice B shows the general solution without applying the initial condition, which is a common error. When solving differential equations with initial conditions, always substitute the initial values into the general solution to determine the specific constant.
A function satisfies $\frac{dy}{dx}=\frac{y}{x}$ for $x>0$ with $y(2)=7$ . Which $y(x)$ is correct?
Answer: $y=\frac{7}{2}x$ This homogeneous differential equation requires applying an initial condition at a non-zero point. The general solution to $\frac{dy}{dx}=\frac{y}{x}$ is $y=Cx$ (found by separating variables: $\frac{dy}{y}=\frac{dx}{x}$ , giving $\ln|y|=\ln|x|+\ln|C|$ ). Using the initial condition $y(2)=7$ : $7=C(2)$ , so $C=\frac{7}{2}$ . Therefore, the particular solution is $y=\frac{7}{2}x$ . Choice B incorrectly assumes $C=7$ , forgetting to solve for $C$ using the given point $(2,7)$ . When the initial condition is given at a point other than $x=0$ , substitute both coordinates to find the constant.
Suppose $\frac{dy}{dx}=\frac{2y}{x}$ and $y(1)=5$ . Which function is the particular solution?
Answer: $y=5x^2$ This problem involves using an initial condition with a homogeneous differential equation. Separating variables in dy/dx = 2y/x gives dy/y = 2dx/x, which integrates to ln|y| = 2ln|x| + C = ln|x²| + C. This yields y = Ax² where A = $e^C$ . Applying the initial condition y(1) = 5: 5 = A(1)² = A·1 = A. Therefore A = 5 and the particular solution is y = 5x². Choice D incorrectly inverts the x² term, perhaps confusing the integration of 1/x. When solving dy/y = k·dx/x, the solution always has the form y = $Cx^k$ , where the initial condition determines C.
A function satisfies $\dfrac{dy}{dx}=\dfrac{y}{x-1}$ with $y(2)=4$ ; find the particular solution $y(x)$ .
Answer: $y=4(x-1)$ This problem requires applying the initial condition to find the particular solution after solving the differential equation using separation of variables. First, separate variables: dy/y = dx/(x - 1), then integrate both sides to get ln|y| = ln|x - 1| + C. Exponentiate to solve for y: y = $e^C$ (x - 1), or y = K (x - 1). Use the initial condition y(2) = 4: 4 = K * 1, so K = 4, yielding y = 4 (x - 1). A tempting distractor is y = 4 (x + 1), which might result from incorrect variable in logarithm. Always plug the initial condition into the general solution to determine the constant C and verify by differentiating back.
A bacteria culture satisfies $\frac{dP}{dt}=0.4P$ with $P(0)=50$ . Which function gives $P(t)$ ?
Answer: $P(t)=50e^{0.4t}$ The skill demonstrated in this problem is applying an initial condition to find the particular solution of a separable differential equation. The general solution to dP/dt = 0.4P is P(t) = C $e^{0.4t}$ , found by separating variables, integrating, and exponentiating. Substituting the initial condition P(0) = 50 gives 50 = C $e^{0}$ , so C = 50. Therefore, the particular solution is P(t) = 50 $e^{0.4t}$ . A tempting distractor is P(t) = C $e^{0.4t}$ , which is the general solution, but it fails because it does not use the initial condition to determine the specific value of C. Always substitute the initial values into the general solution to solve for the arbitrary constant and identify the unique function satisfying both the differential equation and the initial condition.
A tank's salt mass $S$ satisfies $\frac{dS}{dt}=-0.5S$ . If $S(2)=10$ , find $S(t)$ .
Answer: $S(t)=10e^{-0.5(t-2)}$ This problem requires finding a particular solution using an initial condition given at t = 2 rather than t = 0. The general solution to dS/dt = -0.5S is S(t) = $Ce^{-0.5t}$ . To find C, we use S(2) = 10: 10 = $Ce^{-0.5·2}$ = $Ce^{-1}$ . Solving for C gives C = $10e^1$ = 10e. Therefore, S(t) = 10e· $e^{-0.5t}$ = $10e^{1-0.5t}$ = $10e^{-0.5(t-2)}$ . Choice B incorrectly assumes the initial condition is at t = 0, which would give S(t) = $10e^{-0.5t}$ . When the initial condition is given at t = t₀ ≠ 0, the solution often takes the form involving (t - t₀) in the exponent.

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