Intro to Separable Differential Equations & Simple Exponential Models

We start by separating the variables. Multiply both sides by y to get y dy = 3x² dx. Next, integrate both sides. The integral of y with respect to y is (½)y². The integral of 3x² with respect to x is x³. This gives us (½)y² = x³ + C, where C is the constant of integration. Multiply both sides by 2 to get y² = 2x³ + 2C. Since 2C is also a constant, we can rename it as a new constant C. So, y² = 2x³ + C. Finally, take the square root of both sides to solve for y, yielding y = √(2x³ + C).

A separable differential equation can be written in the form dy/dx = f(x)g(y), where the right-hand side is a product of a function of x and a function of y. Option B, dy/dx = xy, fits this form because it can be written as (x)(y), where f(x)=x and g(y)=y. Option A cannot be separated into a product of a function of x and a function of y. Option C involves the argument (x+y) inside the sine function, which cannot be written as a product. Option D has y in the denominator in a way that cannot be factored into a product of a function of x and a function of y.

When money earns interest continually at a rate r, the rate of change of the amount is proportional to the current amount. The constant of proportionality is the interest rate. Here, r = 5% = 0.05. Therefore, the differential equation is dP/dt = 0.05 P. This equation is separable because it can be written as (1/P) dP = 0.05 dt.

First, separate the variables by dividing both sides by y and multiplying by dx: (1/y) dy = 2x dx. Now integrate both sides. The integral of (1/y) with respect to y is ln|y|. The integral of 2x with respect to x is x². So we have ln|y| = x² + C. To solve for y, exponentiate both sides: e^(ln|y|) = ex²+C. This simplifies to |y| = ex²eC. Since eC is a positive constant, we can write it as a new constant C1 > 0. Then y = ± C1ex². The ± and the constant can be combined into a single constant C, which can be any nonzero real number. Often we write y = C ex², where C can be any real number (including zero, but note that if C=0, then y=0 is also a solution, which was lost when we divided by y. However, in the separation process, we assumed y≠0. The constant zero can be included by allowing C=0). So the general solution is y = C ex².

The key step in solving a separable differential equation is to separate the variables. This means rearranging the equation so that one side contains only y and dy, and the other side contains only x and dx. After separation, we integrate both sides to find the solution.

First, separate the variables: (1/y²) dy = cos(x) dx. Integrate both sides. The left side integrates to -1/y. The right side integrates to sin(x). So we have -1/y = sin(x) + C. Multiply both sides by -1 to get 1/y = -sin(x) - C. We can rename -C as a new constant, say C1. So 1/y = -sin(x) + C1. Apply the initial condition y(0)=1, which gives 1/1 = -sin(0) + C1, so 1 = 0 + C1, so C1 = 1. Therefore, 1/y = -sin(x) + 1, or 1/y = 1 - sin(x). Taking reciprocals, y = 1/(1 - sin(x)).

For a separable differential equation dy/dx = f(x)g(y), we first separate the variables by dividing both sides by g(y) (assuming g(y) ≠ 0) to get (1/g(y)) dy/dx = f(x). Then we multiply both sides by dx to obtain (1/g(y)) dy = f(x) dx. Integrating both sides gives the general solution.

Separate variables: y dy = eˣ dx. Integrate both sides: ∫ y dy = ∫ eˣ dx. The left side gives (½)y², and the right side gives eˣ. So (½)y² = eˣ + C. Multiply both sides by 2: y² = 2eˣ + 2C. Since 2C is a constant, rename it as C: y² = 2eˣ + C. Then take the square root: y = ± √(2eˣ + C). Usually we write y = √(2eˣ + C) with the understanding that the constant can be adjusted to include the ±.

A separable differential equation can be written as dy/dx = f(x)g(y). Option A is separable because it is y * sin(x). Option C is separable because it is x² * y². Option D: e^(x+y) = eˣ e^y, so it is separable as eˣ * e^y. Option B cannot be written as a product of a function of x and a function of y; it is a sum, so it is not separable.

Separate variables: dy/(y+1) = 2x dx. Integrate both sides: ∫ dy/(y+1) = ∫ 2x dx. The left side gives ln|y+1|, and the right side gives x². So ln|y+1| = x² + C. Exponentiate: |y+1| = ex²+C = eC ex². Let K = eC (positive constant). Then y+1 = ± K ex². Combine the ± into the constant to get y+1 = C1 ex², where C1 can be any nonzero constant. So y = C1 ex² - 1. Use the initial condition y(0)=0: 0 = C1 e⁰- 1 = C1 - 1, so C1 = 1. Therefore, y = ex² - 1.

In the separation of variables method, we rearrange the differential equation so that all terms containing y and dy are on one side and all terms containing x and dx are on the other side. Then we integrate the left side with respect to y and the right side with respect to x.

Separate variables: y dy = (x² + 1) dx. Integrate both sides: ∫ y dy = ∫ (x² + 1) dx. The left side gives (½)y². The right side gives (⅓)x³ + x. So (½)y² = (⅓)x³ + x + C. Multiply both sides by 2: y² = (⅔)x³ + 2x + 2C. Rename 2C as C: y² = (⅔)x³ + 2x + C. Then y = ± √((⅔)x³ + 2x + C).

The condition for using the method of separation of variables is that we can algebraically manipulate the equation to separate the variables, i.e., write it as (1/g(y)) dy = f(x) dx. Then we can integrate both sides. The functions f(x) and g(y) do not need to be constant; they just need to be integrable.

Separate variables: (1/y) dy = (1/x) dx. Integrate both sides: ∫ (1/y) dy = ∫ (1/x) dx. This gives ln|y| = ln|x| + C. Exponentiate both sides: e^(ln|y|) = e^(ln|x| + C) = e^C * e^(ln|x|) = C1 |x|, where C1 = e^C > 0. So |y| = C1 |x|, which means y = ± C1 x. Let C₂ = ± C1, which can be any nonzero constant. Then y = C₂ x. Use the initial condition y(1)=2: 2 = C₂ * 1, so C₂ = 2. Therefore, y = 2x.

Separate variables: (1/y) dy = 5x⁴ dx. Integrate both sides: ∫ (1/y) dy = ∫ 5x⁴ dx. The left side gives ln|y|, and the right side gives x⁵. So ln|y| = x⁵ + C. Exponentiate: |y| = ex⁵ + C = e^C ex⁵. Let K = eC (positive constant). Then y = ± K ex⁵. The ± and the positive constant K can be combined into a single arbitrary constant C, which can be any nonzero real number. (Note: C = 0 gives the trivial solution y = 0, which was lost when we divided by y, but is also a solution to the original equation.) Thus the general solution is written as y = C ex⁵.