Separable Equations: Modeling, Implicit Solutions & Uniqueness Considerations

To solve a separable differential equation, we need to group all terms involving y with dy and all terms involving x with dx. We start with the given equation dy/dx = y/x. By multiplying both sides by dx, we get dy = (y/x) dx. Next, we divide both sides by y to move the y term to the left side. This results in (1/y) dy = (1/x) dx. Finally, we integrate both sides to set up the solution, which matches the correct option.

We begin by separating the variables. We divide both sides by (y² + 1) and multiply by dx to get 1/(y² + 1) dy = 2x dx. Next, we integrate both sides. The integral of 1/(y² + 1) with respect to y is arctan(y), and the integral of 2x with respect to x is x². Don't forget to add the constant of integration, C, to the right side. This gives us the equation arctan(y) = x² + C. To isolate y, we take the tangent of both sides, resulting in the general solution y = tan(x² + C).

The phrase "rate of change" refers to the derivative with respect to time, which is dP/dt. The problem states that this rate is "proportional to its current population." In mathematical terms, "proportional to P" means we multiply P by a constant, k. Therefore, equating the rate to the proportional amount gives us the differential equation dP/dt = kP. This is the standard model for exponential growth.

First, we use the laws of exponents to rewrite the right side of the equation. The term e^(x-y) can be written as eˣ / e^y. So, the equation becomes dy/dx = eˣ / e^y. To separate the variables, we multiply both sides by e^y and by dx, resulting in e^y dy = eˣ dx. Now we integrate both sides. The integral of e^y dy is simply e^y, and the integral of eˣ dx is eˣ. Adding the constant of integration gives us the implicit general solution e^y = eˣ + C.

We start by separating the variables to get y dy = x dx. Integrating both sides yields (½)y² = (½)x² + C1. Multiplying the entire equation by 2 simplifies this to y² = x² + C, where C is a new constant. To find C, we apply the initial condition y(0) = -4. Plugging these values in gives (-4)² = 0² + C, so 16 = C. The equation is now y² = x² + 16. Solving for y requires taking the square root, which gives y = plus or minus the square root of (x² + 16). Since the initial y-value is negative (-4), we must choose the negative branch. Thus, the particular solution is y = - square root of (x² + 16).

Before separating, we must factor the right side of the equation using grouping. We look at xy + 2x - y - 2. We can factor an x out of the first two terms to get x(y+2), and factor a -1 out of the last two terms to get -1(y+2). This allows us to write the equation as dy/dx = (x-1)(y+2). Now that the expression is a product of functions of x and y, we can separate them. Dividing by (y+2) and multiplying by dx gives us dy/(y+2) = (x-1) dx.

First, separate the variables by dividing by (1+y²) and multiplying by dx, which yields dy/(1+y²) = tan(x) dx. Integrating the left side gives arctan(y). Integrating the right side, the integral of tan(x), results in ln|sec(x)| + C. So, we have arctan(y) = ln|sec(x)| + C. To isolate y, we take the tangent of both sides. This results in the general solution y = tan(ln|sec(x)| + C).

We start by separating the variables. Multiplying by y and dx gives y dy = -x * e^(-x²) dx. We integrate the left side to get (½)y². For the right side, we use u-substitution where u = -x², so du = -2x dx, or (½)du = -x dx. This makes the integral (½)e^u du, which integrates to (½)e^(-x²) + C. Equating the sides, we have (½)y² = (½)e^(-x²) + C. Multiplying by 2 gives y² = e^(-x²) + C. We apply the initial condition (0, 1): 1² = e⁰+ C, which means 1 = 1 + C, so C = 0. The equation becomes y² = e^(-x²). Taking the square root, and noting y is positive at the initial point, we get y = square root of (e^(-x²)).

Continuous interest implies that the rate at which the money grows is proportional to the current amount of money. The rate of change of money with respect to time is dP/dt. The proportionality constant is the interest rate, r. Therefore, the rate of change is equal to the rate times the amount, leading to the equation dP/dt = rP. This is a classic application of separable differential equations in finance.

We separate the variables by dividing by y and multiplying by dx, resulting in (1/y) dy = (ln(x)/x) dx. Integrating the left side gives ln|y|. For the right side, we use u-substitution. Let u = ln(x), so du = (1/x) dx. The integral becomes the integral of u du, which is (½)u². Substituting back, we get (½)(ln x)² + C. Equating the two sides gives the implicit solution ln|y| = (½)(ln x)² + C.

Separate the variables to get dy/((y-1)(y+1)) = dx/x. We use partial fraction decomposition on the left side: 1/((y-1)(y+1)) = (½)/(y-1) - (½)/(y+1). Integrating this yields (½)(ln|y-1| - ln|y+1|) = ln|x| + C1. Multiplying by 2 gives ln|(y-1)/(y+1)| = 2ln|x| + 2C1 = ln(x²) + C₂. Exponentiating both sides leads to (y-1)/(y+1) = Ax² (where A is a constant). Solving for y involves algebraic rearrangement: y - 1 = Ax²(y + 1) -> y - Ax²y = Ax² + 1 -> y(1 - Ax²) = 1 + Ax². Thus, y = (1 + Ax²)/(1 - Ax²).

First, separate the variables: y^(-2) dy = sin(x) dx. Integrating the left side gives -y^(-1) or -1/y. Integrating the right side gives -cos(x) + C. So, -1/y = -cos(x) + C. We can multiply by -1 to get 1/y = cos(x) - C (note that -C is just an arbitrary constant, so we can write it as +K, but keeping the sign consistent helps). Using the initial condition y(0)=1, we get 1/1 = cos(0) - C, which means 1 = 1 - C, so C = 0. The equation is 1/y = cos(x). Taking the reciprocal of both sides gives y = 1/cos(x), which is also sec(x).

Let's look at the separation for each. Option A separates to y dy = x * eˣ dx. The right side, the integral of x * eˣ dx, requires Integration by Parts because it is a product of algebraic (x) and exponential (eˣ) functions. Option B separates to dy/y = x dx (simple power rule). Option C separates to y dy = sin(x) dx (simple trig). Option D separates to dy/y = x^(-2) dx (simple power rule). Therefore, only A requires the advanced technique.

First, factor the numerator and denominator on the right side. The numerator factors to y(x+2). The denominator is a perfect square trinomial, (x+2)². The equation simplifies to dy/dx = y(x+2) / (x+2)², which further simplifies to dy/dx = y / (x+2). Now, separate the variables: (1/y) dy = (1/(x+2)) dx. Integrating both sides yields ln|y| = ln|x+2| + C.

This is a subtle problem involving existence and uniqueness. First, we can see by inspection that the constant function y = 0 is a solution because the derivative of 0 is 0, and the right side 3(0)^(⅔) is also 0. However, if we separate variables by dividing by y^(⅔), we get y^(-2/3) dy = 3 dx. Note that this division is only valid when y ≠ 0, which is why the separation method loses the y = 0 solution. Integrating gives 3y^(⅓) = 3x + C. If y(0)=0, then C=0, so 3y^(⅓) = 3x, which simplifies to y^(⅓) = x, or y = x³. Both y = 0 and y = x³ satisfy the differential equation and the initial condition. This non-uniqueness occurs because ∂f/∂y = 2y^(-⅓) is not continuous at y = 0, violating the conditions of the existence and uniqueness theorem.