Applications of Separable Equations: Cooling, Mixing, Logistic & Trigonometric Models

A differential equation is separable if it can be written in the form dy/dx = g(x)h(y). Option A involves a sum, which cannot be factored into a product. Option B has x and y inside a sine function, which cannot be separated. Option D is a sum of squares, which is not factorable into functions of x and y alone. However, Option C can be rewritten using exponent rules as dy/dx = eˣ * e^y. This is a product of a function of x and a function of y, making it separable.

The "rate of change of temperature" is dT/dt. "Proportional to" implies multiplication by a constant, k. The "difference between its temperature and ambient temperature" is (T - Ta). Putting this together, we get dT/dt = k(T - Ta). Note that k is usually negative in cooling problems to indicate the temperature is decreasing toward the ambient temperature, but the structure of the equation remains the same.

Separation gives dy/(y² - 1) = dx/(x² - 1). Both sides require partial fraction decomposition. The identity is 1/(u² - 1) = (½)/(u-1) - (½)/(u+1). Integrating both sides yields (½)(ln|y-1| - ln|y+1|) = (½)(ln|x-1| - ln|x+1|) + C1. Multiplying by 2 and combining logs gives ln|(y-1)/(y+1)| = ln|(x-1)/(x+1)| + C₂. Exponentiating both sides removes the natural logs, resulting in (y-1)/(y+1) = A * (x-1)/(x+1), where A is a constant.

The rate of change of salt, dy/dt, is (Rate In) - (Rate Out).

Rate In = (concentration in) * (flow rate in) = 0.5 kg/L * 2 L/min = 1 kg/min.

Rate Out = (concentration in tank) * (flow rate out). The concentration in the tank is y/100 (amount y divided by volume 100L). The flow rate out is 2 L/min. So Rate Out = (y/100) * 2 = y/50 kg/min.

Combining these, dy/dt = 1 - y/50.

Separate variables: dy/y = cot(x) dx. Recall that cot(x) = cos(x)/sin(x). Integrating dy/y gives ln|y|. Integrating cot(x) gives ln|sin(x)| + C. So, ln|y| = ln|sin(x)| + C. Exponentiating both sides gives y = A sin(x). Using the initial condition y(π/2) = 2: 2 = A sin(π/2). Since sin(π/2) = 1, we get A = 2. Thus, the solution is y = 2 sin(x).

To separate the variables, we divide both sides by sec(y). Since 1/sec(y) is equal to cos(y), the equation becomes cos(y) dy = cos(x) dx. Now we integrate both sides. The antiderivative of cos(y) with respect to y is sin(y). The antiderivative of cos(x) with respect to x is sin(x). Adding the constant of integration, we arrive at the implicit solution sin(y) = sin(x) + C.

We separate the variables by multiplying both sides by y and dx. This gives us y dy = ((1+x)/x) dx. We can simplify the right side by dividing each term in the numerator by x, resulting in y dy = (1/x + 1) dx. Now we integrate. The left side becomes (½)y². The right side becomes ln|x| + x + C1. To clear the fraction, we multiply the entire equation by 2, yielding y² = 2ln|x| + 2x + C (where C is 2*C1).

Separate the variables to get (1/y) dy = 4x³ dx. Integrating both sides gives ln|y| = x⁴ + C. To isolate y, we exponentiate both sides: |y| = e^(x⁴ + C) = e^C * e^(x⁴). We can rename e^C as a new constant A, so y = Ae^(x⁴). Using the initial condition y(0) = 5, we substitute x=0 and y=5 to get 5 = Ae^0, which implies A = 5. Therefore, the particular solution is y = 5e^(x⁴).

Rewrite the equation using exponent laws: dy/dx = e²ˣ * e^y. Separate variables by dividing by e^y (multiplying by e^(-y)) and multiplying by dx: e^(-y) dy = e²ˣ dx. Integrate the left side: the integral of e^(-y) is -e^(-y). Integrate the right side: the integral of e²ˣ is (½)e²ˣ. Adding the constant gives -e^(-y) = (½)e²ˣ + C.

First, separate the variables: dy/y = dx / (x * ln(x)). The left side integrates to ln|y|. For the right side, we have the integral of 1/(x * ln(x)) dx. If we let u = ln(x), then du = (1/x) dx. The integral becomes integral of (1/u) du, which is ln|u| = ln|ln(x)|. Therefore, the substitution u = ln(x) is the correct step to solve the integral.

Separate variables: dy/(1+y²) = dx/(1+x²). Integrating both sides gives arctan(y) = arctan(x) + K. To solve for y, we take the tangent of both sides: y = tan(arctan(x) + K). Using the tangent addition formula tan(A+B) = (tan A + tan B)/(1 - tan A tan B), we get y = (x + tan(K)) / (1 - x tan(K)). Letting C = tan(K), the solution simplifies to y = (x + C) / (1 - Cx). Note that option B represents the implicit form of the solution, but since the question asks for y expressed explicitly as a function of x, option A is the correct answer.

The standard form of the logistic equation is dP/dt = rP(1 - P/K), where r is the growth rate and K is the carrying capacity. The carrying capacity represents the maximum stable population size where the growth rate becomes zero. In the given equation dP/dt = 0.1P(1 - P/100), the term corresponding to K is 100. Therefore, the carrying capacity is 100.

Separate variables: dy / square root(1 - y²) = x dx. Integrating the left side gives arcsin(y). Integrating the right side gives x²/2 + C. So, arcsin(y) = x²/2 + C. Using the initial condition y(0)=0: arcsin(0) = 0²/2 + C, which means 0 = 0 + C, so C = 0. The equation is arcsin(y) = x²/2. Taking the sine of both sides yields y = sin(x²/2).

Constant solutions occur when the derivative dy/dx is zero for all x. We set the right side of the equation to zero: y² - 4 = 0. Factoring this gives (y - 2)(y + 2) = 0. This yields two solutions: y = 2 and y = -2. If y is constantly 2 or -2, its derivative is 0, and the right side of the differential equation is also 0, satisfying the equation.

We separate the variables by multiplying by (y² + y) and dx. This gives (y² + y) dy = x dx. We integrate the left side using the power rule: the integral of y² is y³/3, and the integral of y is y²/2. We integrate the right side to get x²/2 + C. Equating them gives the implicit general solution y³/3 + y²/2 = x²/2 + C.