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.

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.

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.

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).

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).

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.