Initial Conditions and Separation of Variables - AP Calculus AB

A solution to $\frac{dy}{dx}=\sin x\, y$ satisfies $y(0)=2$ . Which is the particular solution?
Answer: $y=2e^{1-\cos x}$ This problem requires using initial conditions to find the particular solution to a differential equation solved by separation of variables. After separating variables and integrating, we obtain the general solution $y = C e^{-\cos x}$ . We then plug in the initial condition $y(0) = 2$ to find $C = 2 e$ , giving $y = 2 e^{1 - \cos x}$ . This determines the specific function that satisfies both the differential equation and the initial value. A common distractor, such as choice B, fails because it omits the constant adjustment, resulting in $y(0) = 2 e^{-1} \neq 2$ . In general, to apply initial conditions, substitute the given point into the general solution and solve for the constant to get the particular solution.
A solution to $\frac{dy}{dx}=\frac{2y}{1+x}$ satisfies $y(1)=5$ . Which is the particular solution?
Answer: $y=\frac{5}{4}(1+x)^2$ This problem requires using initial conditions to find the particular solution to a differential equation solved by separation of variables. After separating variables and integrating, we obtain the general solution y = C $(1 + x)^2$ . We then plug in the initial condition y(1) = 5 to find C = 5/4. This determines the specific function that satisfies both the differential equation and the initial value. A common distractor, such as choice B, fails because it omits the fractional constant, resulting in y(1) = 20 ≠ 5. In general, to apply initial conditions, substitute the given point into the general solution and solve for the constant to get the particular solution.
A solution to $\frac{dy}{dx}=\frac{3}{y}$ satisfies $y(2)=3$ . Which is the particular solution?
Answer: $y=\sqrt{6x-3}$ This problem requires using initial conditions to find the particular solution to a differential equation solved by separation of variables. After separating variables and integrating, we obtain the general solution $y^2$ = 6x + C. We then plug in the initial condition y(2) = 3 to find C = -3, giving y = √(6x - 3). This determines the specific function that satisfies both the differential equation and the initial value. A common distractor, such as choice B, fails because it uses a positive constant incorrectly, resulting in y(2) = √15 ≠ 3. In general, to apply initial conditions, substitute the given point into the general solution and solve for the constant to get the particular solution.
A solution to $\frac{dy}{dx}=\frac{x}{y}$ satisfies $y(0)=3$ . Which is the particular solution?
Answer: $y=\sqrt{x^2+9}$ This problem requires using initial conditions to find the particular solution to a differential equation solved by separation of variables. After separating variables and integrating, we obtain the general solution $y^2 = x^2 + C$ . We then plug in the initial condition $y(0) = 3$ to find C = 9, giving $y = \sqrt{x^2 + 9}$ . This determines the specific function that satisfies both the differential equation and the initial value. A common distractor, such as choice B, fails because it omits the constant, resulting in $y(0) = 0 \neq 3$ . In general, to apply initial conditions, substitute the given point into the general solution and solve for the constant to get the particular solution.
A solution to $\frac{dy}{dx}=\frac{2x+3}{y}$ satisfies $y(0)=2$ . Which is the particular solution?
Answer: $y=\sqrt{2x^2+6x+4}$ This problem requires using initial conditions to find the particular solution to a differential equation solved by separation of variables. After separating variables and integrating, we obtain the general solution $y^2$ = $2x^2$ + 6x + C. We then plug in the initial condition y(0) = 2 to find C = 4, giving y = √( $2x^2$ + 6x + 4). This determines the specific function that satisfies both the differential equation and the initial value. A common distractor, such as choice B, fails because it omits the constant, resulting in y(0) = 0 ≠ 2. In general, to apply initial conditions, substitute the given point into the general solution and solve for the constant to get the particular solution.
A solution to $\frac{dy}{dx}=-\frac{2y}{x}$ satisfies $y(2)=1$ . Which is the particular solution?
Answer: $y=\frac{4}{x^2}$ This problem requires using initial conditions to find the particular solution to a differential equation solved by separation of variables. After separating variables and integrating, we obtain the general solution y = C / $x^2$ . We then plug in the initial condition y(2) = 1 to find C = 4, giving y = 4 / $x^2$ . This determines the specific function that satisfies both the differential equation and the initial value. A common distractor, such as choice B, fails because it halves the constant incorrectly, resulting in y(2) = 2/4 = 0.5 ≠ 1. In general, to apply initial conditions, substitute the given point into the general solution and solve for the constant to get the particular solution.
A bacteria culture satisfies $\frac{dB}{dt}=3B$ with $B(1)=2$ . Which is the particular solution $B(t)$ ?
Answer: $B(t)=2e^{3(t-1)}$ This problem requires using initial conditions to find the particular solution to a differential equation solved by separation of variables. After separating variables and integrating, we obtain the general solution B(t) = C $e^{3t}$ . We then plug in the initial condition B(1) = 2 to find C = 2 $e^{-3}$ , which gives B(t) = 2 $e^{3(t-1)}$ . This determines the specific function that satisfies both the differential equation and the initial value. A common distractor, such as choice A, fails because it uses the initial condition at t=0 instead of t=1, resulting in B(1) = 2 $e^{3}$ ≠ 2. In general, to apply initial conditions, substitute the given point into the general solution and solve for the constant to get the particular solution.
A solution to $\frac{dy}{dx}=xy$ passes through $(0,5)$ . Which is the particular solution?
Answer: $y=5e^{x^2/2}$ This problem requires using initial conditions to find the particular solution to a differential equation solved by separation of variables. After separating variables and integrating, we obtain the general solution y = C $e^{x^2/2}$ . We then plug in the initial condition y(0) = 5 to find C = 5. This determines the specific function that satisfies both the differential equation and the initial value. A common distractor, such as choice B, fails because it omits the constant from the initial condition, resulting in y(0) = 1 instead of 5. In general, to apply initial conditions, substitute the given point into the general solution and solve for the constant to get the particular solution.
A solution to $\frac{dy}{dx}=(2-x)y$ satisfies $y(1)=3$ . Which is the particular solution?
Answer: $y=3e^{2x-x^2/2-3/2}$ This problem requires using initial conditions to find the particular solution to a differential equation solved by separation of variables. After separating variables and integrating, we obtain the general solution y = C $e^{2x - x^2/2}$ . We then plug in the initial condition y(1) = 3 to find C = 3 $e^{-3/2}$ , giving y = 3 $e^{2x - x^2/2 - 3/2}$ . This determines the specific function that satisfies both the differential equation and the initial value. A common distractor, such as choice B, fails because it ignores the adjustment for the initial condition at x=1. In general, to apply initial conditions, substitute the given point into the general solution and solve for the constant to get the particular solution.
A solution to $\frac{dy}{dx}=\frac{2y}{x}$ satisfies $y(1)=4$ . Which is the particular solution?
Answer: $y=4x^2$ This problem requires using initial conditions to find the particular solution to a differential equation solved by separation of variables. After separating variables and integrating, we obtain the general solution y = C $x^2$ . We then plug in the initial condition y(1) = 4 to find C = 4. This determines the specific function that satisfies both the differential equation and the initial value. A common distractor, such as choice D, fails because it halves the coefficient incorrectly, resulting in y(1) = 2 ≠ 4. In general, to apply initial conditions, substitute the given point into the general solution and solve for the constant to get the particular solution.
A solution to $\frac{dy}{dx}=\frac{y}{x\ln x}$ satisfies $y(e)=4$ . Which is the particular solution?
Answer: $y=4\ln x$ This problem requires using initial conditions to find the particular solution to a differential equation solved by separation of variables. After separating variables and integrating, we obtain the general solution y = C ln x. We then plug in the initial condition y(e) = 4 to find C = 4. This determines the specific function that satisfies both the differential equation and the initial value. A common distractor, such as choice B, fails because it omits the constant, resulting in y(e) = 1 ≠ 4. In general, to apply initial conditions, substitute the given point into the general solution and solve for the constant to get the particular solution.
A solution to $\frac{dy}{dx}=\frac{1+x}{y}$ satisfies $y(0)=1$ . Which is the particular solution?
Answer: $y=\sqrt{x^2+2x+1}$ This problem requires using initial conditions to find the particular solution to a differential equation solved by separation of variables. After separating variables and integrating, we obtain the general solution $y^2$ = $x^2$ + 2x + C. We then plug in the initial condition y(0) = 1 to find C = 1, giving y = √( $x^2$ + 2x + 1). This determines the specific function that satisfies both the differential equation and the initial value. A common distractor, such as choice B, fails because it omits the constant, resulting in y(0) = 0 ≠ 1. In general, to apply initial conditions, substitute the given point into the general solution and solve for the constant to get the particular solution.
A solution to $\frac{dy}{dx}=\frac{1}{y}$ satisfies $y(0)=2$ . Which is the particular solution?
Answer: $y=\sqrt{2x+4}$ This problem requires using initial conditions to find the particular solution to a differential equation solved by separation of variables. After separating variables and integrating, we obtain the general solution $y^2$ = 2x + C. We then plug in the initial condition y(0) = 2 to find C = 4, giving y = √(2x + 4). This determines the specific function that satisfies both the differential equation and the initial value. A common distractor, such as choice B, fails because it omits the constant from the initial condition, resulting in y(0) = 0 ≠ 2. In general, to apply initial conditions, substitute the given point into the general solution and solve for the constant to get the particular solution.
A curve satisfies $\frac{dy}{dx}=\frac{4x}{y^{3}}$ with $y(0)=1$ . Which particular solution is $y(x)$ ?
Answer: $y^{4}=8x^{2}+1$ The problem requires using the initial condition to find the particular solution after separating variables in the differential equation. By separating variables and integrating, we obtain the general solution $y^4$ = $8x^2$ + C. We then substitute the initial values x=0 and y=1 into the general solution to solve for C=1. This determines the specific value of C that makes the solution pass through the given point, yielding $y^4$ = $8x^2$ +1. A tempting distractor might be choice C, $y^4$ = $16x^2$ +1, which fails because it doubles the coefficient incorrectly during integration. To apply initial conditions effectively, always plug in the given values immediately after finding the general solution and solve for the constant carefully.
A curve satisfies $\frac{dy}{dx}=\frac{1-y}{1+x}$ with $y(0)=2$ . Which particular solution gives $y(x)$ ?
Answer: $y=1+\dfrac{1}{1+x}$ The problem requires using the initial condition to find the particular solution after separating variables in the differential equation. By separating variables and integrating, we obtain the general solution y = 1 + K/(1+x). We then substitute the initial values x=0 and y=2 into the general solution to solve for K=1. This determines the specific value of K that makes the solution pass through the given point, yielding y=1+1/(1+x). A tempting distractor might be choice B, y=1+2/(1+x), which fails because it uses an incorrect sign or factor in solving for K. To apply initial conditions effectively, always plug in the given values immediately after finding the general solution and solve for the constant carefully.
A solution to $\frac{dy}{dx}=\frac{y}{x}$ satisfies $y(2)=6$ . Which is the particular solution?
Answer: $y=3x$ This problem requires using initial conditions to find the particular solution to a differential equation solved by separation of variables. After separating variables and integrating, we obtain the general solution y = C x. We then plug in the initial condition y(2) = 6 to find C = 3. This determines the specific function that satisfies both the differential equation and the initial value. A common distractor, such as choice B, fails because it incorrectly doubles the constant, resulting in y(2) = 12 ≠ 6. In general, to apply initial conditions, substitute the given point into the general solution and solve for the constant to get the particular solution.
A solution to $\frac{dy}{dx}=e^x y$ satisfies $y(0)=3$ . Which is the particular solution?
Answer: $y=3e^{e^x-1}$ This problem requires using initial conditions to find the particular solution to a differential equation solved by separation of variables. After separating variables and integrating, we obtain the general solution y = C $e^{e^x}$ . We then plug in the initial condition y(0) = 3 to find C = 3 $e^{-1}$ , giving y = 3 $e^{e^x - 1}$ . This determines the specific function that satisfies both the differential equation and the initial value. A common distractor, such as choice B, fails because it ignores the constant adjustment, resulting in y(0) = 3 e ≠ 3. In general, to apply initial conditions, substitute the given point into the general solution and solve for the constant to get the particular solution.
A solution to $\frac{dy}{dx}=2xy$ satisfies $y(1)=3$ . Which is the particular solution?
Answer: $y=3e^{x^2-1}$ This problem requires using initial conditions to find the particular solution to a differential equation solved by separation of variables. After separating variables and integrating, we obtain the general solution y = C $e^{x^2}$ . We then plug in the initial condition y(1) = 3 to find C = 3 $e^{-1}$ , which gives y = 3 $e^{x^2 - 1}$ . This determines the specific function that satisfies both the differential equation and the initial value. A common distractor, such as choice B, fails because it ignores the adjustment from the initial condition, resulting in y(1) = 3 e ≠ 3. In general, to apply initial conditions, substitute the given point into the general solution and solve for the constant to get the particular solution.
A solution to $\frac{dy}{dx}=(1+x)y$ satisfies $y(0)=4$ . Which is the particular solution?
Answer: $y=4e^{x+x^2/2}$ This problem requires using initial conditions to find the particular solution to a differential equation solved by separation of variables. After separating variables and integrating, we obtain the general solution y = C $e^{x + x^2/2}$ . We then plug in the initial condition y(0) = 4 to find C = 4. This determines the specific function that satisfies both the differential equation and the initial value. A common distractor, such as choice B, fails because it omits the full exponential term from the integration. In general, to apply initial conditions, substitute the given point into the general solution and solve for the constant to get the particular solution.
A solution to $\frac{dy}{dx}=\frac{2x}{y}$ satisfies $y(1)=2$ . Which is the particular solution?
Answer: $y=\sqrt{2x^2+2}$ This problem requires using initial conditions to find the particular solution to a differential equation solved by separation of variables. After separating variables and integrating, we obtain the general solution $y^2$ = $2x^2$ + C. We then plug in the initial condition y(1) = 2 to find C = 2, giving y = √( $2x^2$ + 2). This determines the specific function that satisfies both the differential equation and the initial value. A common distractor, such as choice B, fails because it uses a negative constant, leading to imaginary values at x=1. In general, to apply initial conditions, substitute the given point into the general solution and solve for the constant to get the particular solution.

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