Separable Equations With Powers, Roots & Proportional Growth/Decay

1) Solve: dy/dx = 4x³ y².

Y = 1 / (C - x⁴)

Y = 1 / (C + x⁴)

Y = C - x⁴

Y = C + x⁴

Separate variables: (1/y²) dy = 4x³ dx. Integrate both sides: ∫ y^(-2) dy = ∫ 4x³ dx. The left side gives -1/y. The right side gives x⁴. So -1/y = x⁴ + C. Multiply by -1: 1/y = -x⁴ - C. Let C1 = -C, then 1/y = C1 - x⁴. So y = 1/(C1 - x⁴). Rename C1 as C.

Explanation

Separate variables: (1/y²) dy = 4x³ dx. Integrate both sides: ∫ y^(-2) dy = ∫ 4x³ dx. The left side gives -1/y. The right side gives x⁴. So -1/y = x⁴ + C. Multiply by -1: 1/y = -x⁴ - C. Let C1 = -C, then 1/y = C1 - x⁴. So y = 1/(C1 - x⁴). Rename C1 as C.

2) Which of the following differential equations is separable?

Dy/dx = x² + y

Dy/dx = sin(xy)

Dy/dx = y ln(x)

Dy/dx = x/y + y/x

A separable equation can be written as dy/dx = f(x)g(y). Option C can be written as (ln(x)) * y, so it is separable. Option A is not separable because it is a sum. Option B has the product xy inside the sine function, which cannot be separated into a product of a function of x and a function of y. Option D is a sum of two terms that cannot be written as a single product.

Explanation

A separable equation can be written as dy/dx = f(x)g(y). Option C can be written as (ln(x)) * y, so it is separable. Option A is not separable because it is a sum. Option B has the product xy inside the sine function, which cannot be separated into a product of a function of x and a function of y. Option D is a sum of two terms that cannot be written as a single product.

3) Solve: dy/dx = -x/(y e^(x²)) with y(0) = 1.

Y = 1+√(e^(-x²))

Y = √(1 - e^(-x²))

Y = √(1 + e^(-x²))

Y = e^(-x²/2)

Separate variables: y dy = -x e^(-x²) dx. Integrate both sides: ∫ y dy = ∫ -x e^(-x²) dx. The left side gives (½)y². For the right side, use substitution: let u = -x², then du = -2x dx, so -x dx = (½) du. The integral becomes (½) ∫ e^u du = (½) e^u = (½) e^(-x²). So (½)y² = (½) e^(-x²) + C. Multiply by 2: y² = e^(-x²) + 2C. Use y(0)=1: 1 = 1 + 2C => C=0. So y² = e^(-x²), and since y(0)=1>0, y = √(e^(-x²)) = e^(-x²/2).

Explanation

Separate variables: y dy = -x e^(-x²) dx. Integrate both sides: ∫ y dy = ∫ -x e^(-x²) dx. The left side gives (½)y². For the right side, use substitution: let u = -x², then du = -2x dx, so -x dx = (½) du. The integral becomes (½) ∫ e^u du = (½) e^u = (½) e^(-x²). So (½)y² = (½) e^(-x²) + C. Multiply by 2: y² = e^(-x²) + 2C. Use y(0)=1: 1 = 1 + 2C => C=0. So y² = e^(-x²), and since y(0)=1>0, y = √(e^(-x²)) = e^(-x²/2).

4) A population P(t) grows at a rate proportional to the square root of the population. Which separable differential equation models this?

DP/dt = k √(P)

DP/dt = k P

DP/dt = k / P

DP/dt = k t

The rate of change is proportional to the square root of the population. So dP/dt = k √(P), where k is the constant of proportionality. This is separable because it can be written as (1/√(P)) dP = k dt.

Explanation

The rate of change is proportional to the square root of the population. So dP/dt = k √(P), where k is the constant of proportionality. This is separable because it can be written as (1/√(P)) dP = k dt.

5) Solve: dy/dx = 3x² y².

Y = 1 / (C - x³)

Y = 1 / (C + x³)

Y = C - x³

Y = C + x³

Separate variables: (1/y²) dy = 3x² dx. Integrate both sides: ∫ y^(-2) dy = ∫ 3x² dx. The left side gives -1/y. The right side gives x³. So -1/y = x³ + C. Multiply both sides by -1: 1/y = -x³ - C. Let C1 = -C, then 1/y = C1 - x³. So y = 1/(C1 - x³). Rename C1 as C: y = 1/(C - x³).

Explanation

Separate variables: (1/y²) dy = 3x² dx. Integrate both sides: ∫ y^(-2) dy = ∫ 3x² dx. The left side gives -1/y. The right side gives x³. So -1/y = x³ + C. Multiply both sides by -1: 1/y = -x³ - C. Let C1 = -C, then 1/y = C1 - x³. So y = 1/(C1 - x³). Rename C1 as C: y = 1/(C - x³).

6) When solving dy/dx = f(x)g(y), we must consider:

The case when g(y) = 0.

The case when f(x) = 0.

Both A and B.

Neither A nor B.

When separating variables, we divide by g(y). This is only valid if g(y) ≠ 0. So we must consider the case when g(y)=0 separately, because that may give constant solutions that are lost during the division.

Explanation

When separating variables, we divide by g(y). This is only valid if g(y) ≠ 0. So we must consider the case when g(y)=0 separately, because that may give constant solutions that are lost during the division.

7) The solution to dy/dx = k y, where k is a constant, is:

Y = C ekx

Y = C eˣ

Y = C xk

Y = C + kx

This is the standard exponential growth/decay equation. Separate variables: (1/y) dy = k dx. Integrate: ln|y| = kx + C. Exponentiate: y = ekx+C = eC ekx = C ekx, where C is a constant.

Explanation

This is the standard exponential growth/decay equation. Separate variables: (1/y) dy = k dx. Integrate: ln|y| = kx + C. Exponentiate: y = ekx+C = eC ekx = C ekx, where C is a constant.

8) Solve: dy/dx = (1 + y²) / (1 + x²).

Arctan(y) = arctan(x) + C

Y = x + C

Y = (1+x²)/(1+y²) + C

Y = arctan(x) + C

Separate variables: dy/(1+y²) = dx/(1+x²). Integrate both sides: ∫ dy/(1+y²) = ∫ dx/(1+x²). The integral of 1/(1+u²) is arctan(u). So we get arctan(y) = arctan(x) + C.

Explanation

Separate variables: dy/(1+y²) = dx/(1+x²). Integrate both sides: ∫ dy/(1+y²) = ∫ dx/(1+x²). The integral of 1/(1+u²) is arctan(u). So we get arctan(y) = arctan(x) + C.

9) Which of the following is a necessary condition for using separation of variables?

The differential equation must be linear.

The differential equation must be homogeneous.

The differential equation must be of the form dy/dx = f(x)g(y).

The differential equation must have constant coefficients.

Separation of variables applies specifically to equations that can be written as dy/dx = f(x)g(y), or more generally, to equations that can be manipulated so that all x terms are on one side and all y terms are on the other.

Explanation

Separation of variables applies specifically to equations that can be written as dy/dx = f(x)g(y), or more generally, to equations that can be manipulated so that all x terms are on one side and all y terms are on the other.

10) Find the particular solution to dy/dx = y cos(x) with y(0) = 3.

Y = 3 esin(x)

Y = 3 ecos(x)

Y = 3 sin(x)

Y = 3 cos(x)

Separate variables: (1/y) dy = cos(x) dx. Integrate: ln|y| = sin(x) + C. Exponentiate: |y| = esin(x)+C = eC esin(x). Let K = eC, then y = K esin(x). Use y(0)=3: 3 = K e^(sin(0)) = K e⁰= K, so K=3. Therefore, y = 3 esin(x).

Explanation

Separate variables: (1/y) dy = cos(x) dx. Integrate: ln|y| = sin(x) + C. Exponentiate: |y| = esin(x)+C = eC esin(x). Let K = eC, then y = K esin(x). Use y(0)=3: 3 = K e^(sin(0)) = K e⁰= K, so K=3. Therefore, y = 3 esin(x).

11) The general solution to dy/dx = 2xy + 4x can be found by first:

Dividing both sides by x.

Factoring the right side as 2x(y+2).

Integrating both sides with respect to x.

Using the quadratic formula.

The equation is dy/dx = 2xy + 4x. Factor out 2x: dy/dx = 2x(y+2). This is now in separable form: dy/(y+2) = 2x dx. So the first step is factoring.

Explanation

The equation is dy/dx = 2xy + 4x. Factor out 2x: dy/dx = 2x(y+2). This is now in separable form: dy/(y+2) = 2x dx. So the first step is factoring.

12) Solve: dy/dx = x / (y (1 + x²)).

Y² = ln(1+x²) + C

Y² = (½) ln(1+x²) + C

Y² = ln(1+x²) + 2C

Y² = 2 ln(1+x²) + C

Separate variables: y dy = x/(1+x²) dx. Integrate both sides: ∫ y dy = ∫ x/(1+x²) dx. Left side: (½)y². Right side: use substitution u=1+x², du=2x dx, so x dx = du/2. Then ∫ (1/u) (du/2) = (½) ∫ du/u = (½) ln|u| = (½) ln(1+x²). So (½)y² = (½) ln(1+x²) + C. Multiply by 2: y² = ln(1+x²) + 2C. Rename 2C as C: y² = ln(1+x²) + C.

Explanation

Separate variables: y dy = x/(1+x²) dx. Integrate both sides: ∫ y dy = ∫ x/(1+x²) dx. Left side: (½)y². Right side: use substitution u=1+x², du=2x dx, so x dx = du/2. Then ∫ (1/u) (du/2) = (½) ∫ du/u = (½) ln|u| = (½) ln(1+x²). So (½)y² = (½) ln(1+x²) + C. Multiply by 2: y² = ln(1+x²) + 2C. Rename 2C as C: y² = ln(1+x²) + C.

13) A separable differential equation is so named because:

The variables can be separated on opposite sides of the equation.

The solution can be separated into multiple parts.

It involves separate integrals.

It separates continuous and discontinuous functions.

The term "separable" refers to the ability to separate the variables x and y to different sides of the equation, each side involving only one variable.

Explanation

The term "separable" refers to the ability to separate the variables x and y to different sides of the equation, each side involving only one variable.

14) Solve: dy/dx = -y/x² with y(1) = e.

Y = e^(1 - 1/x)

Y = e^(1/x)

Y = e^(1 - x)

Y = e^(x - 1)

Separate variables: (1/y) dy = (-1/x²) dx. Integrate both sides: ∫ (1/y) dy = ∫ -1/x² dx. The left side gives ln|y|. The right side gives 1/x (since the integral of -1/x² is 1/x). So ln|y| = 1/x + C. Exponentiate: |y| = e^(1/x + C) = e^C e^(1/x). Let K = e^C, then y = K e^(1/x). Use the initial condition y(1)=e: e = K e^(1/1) = K e, so K = 1. Therefore, y = e^(1/x).

Explanation

Separate variables: (1/y) dy = (-1/x²) dx. Integrate both sides: ∫ (1/y) dy = ∫ -1/x² dx. The left side gives ln|y|. The right side gives 1/x (since the integral of -1/x² is 1/x). So ln|y| = 1/x + C. Exponentiate: |y| = e^(1/x + C) = e^C e^(1/x). Let K = e^C, then y = K e^(1/x). Use the initial condition y(1)=e: e = K e^(1/1) = K e, so K = 1. Therefore, y = e^(1/x).

15) The method of separation of variables always results in an explicit solution where y is isolated as a function of x (y = f(x)).

True

False

Integrating both sides of a separable equation often yields an implicit solution, where y is mixed with x or contained within non-linear functions (e.g., y² + e^y = x² + C). While it is sometimes possible to solve for y algebraically to get an explicit solution, it is not always possible or necessary to do so.

Explanation

Integrating both sides of a separable equation often yields an implicit solution, where y is mixed with x or contained within non-linear functions (e.g., y² + e^y = x² + C). While it is sometimes possible to solve for y algebraically to get an explicit solution, it is not always possible or necessary to do so.

Separable Equations with Powers, Roots & Proportional Growth/Decay