Conceptual Separable Equations & Basic Initial Value Problems

1) Which of the following is separable?

Dy/dx = x² + 2xy + y²

Dy/dx = eˣ sin(y)

Dy/dx = ln(x+y)

Dy/dx = (x+y)/x

Option B can be written as eˣ * sin(y), which is a product of a function of x and a function of y. The others cannot be written in that form.

Explanation

Option B can be written as eˣ * sin(y), which is a product of a function of x and a function of y. The others cannot be written in that form.

2) A first-order differential equation is considered separable if and only if it can be written in the form dy/dx = g(x) + h(y).

True

False

A differential equation is separable if the derivative dy/dx can be factored into a product of a function of x and a function of y, such as dy/dx = g(x)h(y). If the terms are separated by addition, like g(x) + h(y), the variables usually cannot be separated to opposite sides of the equation through multiplication or division.

Explanation

A differential equation is separable if the derivative dy/dx can be factored into a product of a function of x and a function of y, such as dy/dx = g(x)h(y). If the terms are separated by addition, like g(x) + h(y), the variables usually cannot be separated to opposite sides of the equation through multiplication or division.

3) If the rate of change of a quantity Q is proportional to Q itself, the differential equation is:

DQ/dt = kQ

DQ/dt = k/t

DQ/dt = kQ²

DQ/dt = k

Proportional to Q means dQ/dt = kQ for some constant k.

Explanation

Proportional to Q means dQ/dt = kQ for some constant k.

4) Solve: dy/dx = y sin(x) with y(0) = 2.

Y = 2 e^(-cos(x))

Y = 2 e^(1-cos(x))

Y = 2 e^(cos(x))

Y = 2 e^(-cos(x)+1)

Separate variables: (1/y) dy = sin(x) dx. Integrate: ln|y| = -cos(x) + C. Exponentiate: |y| = e^(-cos(x)+C) = e^C e^(-cos(x)). Let K = e^C, then y = K e^(-cos(x)). Use y(0)=2: 2 = K e^(-cos(0)) = K e^(-1) = K/e, so K = 2e. Then y = 2e e^(-cos(x)) = 2 e^(1-cos(x)).

Explanation

Separate variables: (1/y) dy = sin(x) dx. Integrate: ln|y| = -cos(x) + C. Exponentiate: |y| = e^(-cos(x)+C) = e^C e^(-cos(x)). Let K = e^C, then y = K e^(-cos(x)). Use y(0)=2: 2 = K e^(-cos(0)) = K e^(-1) = K/e, so K = 2e. Then y = 2e e^(-cos(x)) = 2 e^(1-cos(x)).

5) The constant solutions of dy/dx = y(1-y) are:

Y = 0 and y = 1

Y = 0 only

Y = 1 only

There are no constant solutions.

Constant solutions occur when dy/dx = 0, so set y(1-y)=0. This gives y=0 or y=1.

Explanation

Constant solutions occur when dy/dx = 0, so set y(1-y)=0. This gives y=0 or y=1.

6) Solve: dy/dx = (1+x) / y with y(0) = 3.

Y = √(x² + 2x + 9)

Y = √(x² + 2x + 6)

Y = √(2x + x² + 9)

Y = √(2x + x² + 6)

Separate variables: y dy = (1+x) dx. Integrate: ∫ y dy = ∫ (1+x) dx. Left side: (½)y². Right side: x + (½)x². So (½)y² = x + (½)x² + C. Multiply by 2: y² = 2x + x² + 2C. Use y(0)=3: 9 = 0 + 0 + 2C => 2C=9 => C=9/2. Then y² = 2x + x² + 9. So y = √(2x + x² + 9) (since y(0)=3>0, take positive root).

Explanation

Separate variables: y dy = (1+x) dx. Integrate: ∫ y dy = ∫ (1+x) dx. Left side: (½)y². Right side: x + (½)x². So (½)y² = x + (½)x² + C. Multiply by 2: y² = 2x + x² + 2C. Use y(0)=3: 9 = 0 + 0 + 2C => 2C=9 => C=9/2. Then y² = 2x + x² + 9. So y = √(2x + x² + 9) (since y(0)=3>0, take positive root).

7) Separation of variables is applicable to:

All first order differential equations.

Only linear differential equations.

Equations of the form dy/dx = f(x) + g(y).

Equations where the variables can be separated.

The method applies when we can manipulate the equation to have all x and dx on one side and all y and dy on the other.

Explanation

The method applies when we can manipulate the equation to have all x and dx on one side and all y and dy on the other.

8) The differential equation dy/dx = x/y is separable because:

It can be written as y dy = x dx.

It is linear.

It has constant coefficients.

It can be integrated directly.

Multiplying both sides by y and dx gives y dy = x dx, which separates the variables.

Explanation

Multiplying both sides by y and dx gives y dy = x dx, which separates the variables.

9) Find the particular solution to dy/dx = 2y / x with y(1) = 4.

Y = 4x²

Y = 4x

Y = 4/x

Y = 4/x²

Separate variables: (1/y) dy = (2/x) dx. Integrate: ln|y| = 2 ln|x| + C = ln|x²| + C. Exponentiate: |y| = e^C |x²|. Let K = e^C, then y = K x². Use y(1)=4: 4 = K * 1 => K=4. So y = 4x².

Explanation

Separate variables: (1/y) dy = (2/x) dx. Integrate: ln|y| = 2 ln|x| + C = ln|x²| + C. Exponentiate: |y| = e^C |x²|. Let K = e^C, then y = K x². Use y(1)=4: 4 = K * 1 => K=4. So y = 4x².

10) When we separate variables, we:

Integrate both sides with respect to x.

Differentiate both sides with respect to y.

Move all terms involving y to one side and all terms involving x to the other.

Solve for y first.

Separating variables means rearranging the equation so that one side contains only y and dy, and the other side contains only x and dx.

Explanation

Separating variables means rearranging the equation so that one side contains only y and dy, and the other side contains only x and dx.

11) Solve: dy/dx = (y+1)(x-2).

Ln|y+1| = (½)x² + 2x + C

Ln|y+1| = (½)x² - 2x + C

Y+1 = C e^((½)x² - 2x)

Both B and C

Separate variables: dy/(y+1) = (x-2) dx. Integrate both sides: ∫ dy/(y+1) = ∫ (x-2) dx. The left side gives ln|y+1|. The right side gives (½)x² - 2x. So ln|y+1| = (½)x² - 2x + C. Exponentiate both sides: |y+1| = e^((½)x² - 2x + C) = e^C e^((½)x² - 2x). Let K = e^C, then y+1 = K e^((½)x² - 2x). So both the logarithmic form and the exponential form are valid representations of the solution. Therefore, both B and C are correct, with C being the exponentiated version.

Explanation

Separate variables: dy/(y+1) = (x-2) dx. Integrate both sides: ∫ dy/(y+1) = ∫ (x-2) dx. The left side gives ln|y+1|. The right side gives (½)x² - 2x. So ln|y+1| = (½)x² - 2x + C. Exponentiate both sides: |y+1| = e^((½)x² - 2x + C) = e^C e^((½)x² - 2x). Let K = e^C, then y+1 = K e^((½)x² - 2x). So both the logarithmic form and the exponential form are valid representations of the solution. Therefore, both B and C are correct, with C being the exponentiated version.

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

Y = kx + C

Y = C e^(kx)

Y = C x^k

Y = k eˣ

Integrating both sides with respect to x gives y = kx + C.

Explanation

Integrating both sides with respect to x gives y = kx + C.

13) Solve: dy/dx = y / (x ln(x)) with y(e) = 1.

Y = ln(ln(x))

Y = ln(x)

Y = ln(x)/e

Y = ln(ln(x)) + 1

Separate variables: (1/y) dy = (1/(x ln(x))) dx. Integrate both sides: ∫ (1/y) dy = ∫ (1/(x ln(x))) dx. The left side gives ln|y|. For the right side, use substitution u = ln(x), du = (1/x) dx. Then ∫ (1/u) du = ln|u| = ln|ln(x)|. So ln|y| = ln|ln(x)| + C. Exponentiate: |y| = e^C |ln(x)|. Let K = e^C, then y = K ln(x). Use y(e)=1: 1 = K ln(e) = K * 1, so K=1. Therefore, y = ln(x).

Explanation

Separate variables: (1/y) dy = (1/(x ln(x))) dx. Integrate both sides: ∫ (1/y) dy = ∫ (1/(x ln(x))) dx. The left side gives ln|y|. For the right side, use substitution u = ln(x), du = (1/x) dx. Then ∫ (1/u) du = ln|u| = ln|ln(x)|. So ln|y| = ln|ln(x)| + C. Exponentiate: |y| = e^C |ln(x)|. Let K = e^C, then y = K ln(x). Use y(e)=1: 1 = K ln(e) = K * 1, so K=1. Therefore, y = ln(x).

14) Solve the separable differential equation: dy/dx = x²y.

Y = C e^(x³/3)

Y = C e^(x³)

Y = C e^(x²/2)

Y = C x³

We start by separating the variables. Divide both sides by y and multiply both sides by dx to get (1/y) dy = x² dx. Next, integrate both sides. The integral of (1/y) with respect to y is ln|y|. The integral of x² with respect to x is (⅓)x³. This gives us ln|y| = (⅓)x³ + C, where C is the constant of integration. To solve for y, exponentiate both sides: e^(ln|y|) = e^((⅓)x³ + C). This simplifies to |y| = e^C * e^((⅓)x³). Since e^C is a positive constant, we can combine it with the ± from the absolute value to form a new constant, which we can call C. Therefore, the general solution is y = C e^(x³/3).

Explanation

We start by separating the variables. Divide both sides by y and multiply both sides by dx to get (1/y) dy = x² dx. Next, integrate both sides. The integral of (1/y) with respect to y is ln|y|. The integral of x² with respect to x is (⅓)x³. This gives us ln|y| = (⅓)x³ + C, where C is the constant of integration. To solve for y, exponentiate both sides: e^(ln|y|) = e^((⅓)x³ + C). This simplifies to |y| = e^C * e^((⅓)x³). Since e^C is a positive constant, we can combine it with the ± from the absolute value to form a new constant, which we can call C. Therefore, the general solution is y = C e^(x³/3).

15) Solve the differential equation: dy/dx = (y² + 1) / x, with the initial condition y(1) = 0.

Arctan(y) = ln|x|

Y = tan(ln|x|)

Y = tan(ln|x| + π/4)

Y = tan(ln|x|) - 1

First, separate the variables. Multiply both sides by dx and divide both sides by (y² + 1) to get dy/(y² + 1) = (1/x) dx. Now, integrate both sides. The integral of 1/(y² + 1) with respect to y is arctan(y). The integral of 1/x with respect to x is ln|x|. This gives us arctan(y) = ln|x| + C. Apply the initial condition y(1)=0: arctan(0) = ln|1| + C, which simplifies to 0 = 0 + C, so C = 0. Therefore, arctan(y) = ln|x|. To solve for y, take the tangent of both sides: y = tan(ln|x|).

Explanation

First, separate the variables. Multiply both sides by dx and divide both sides by (y² + 1) to get dy/(y² + 1) = (1/x) dx. Now, integrate both sides. The integral of 1/(y² + 1) with respect to y is arctan(y). The integral of 1/x with respect to x is ln|x|. This gives us arctan(y) = ln|x| + C. Apply the initial condition y(1)=0: arctan(0) = ln|1| + C, which simplifies to 0 = 0 + C, so C = 0. Therefore, arctan(y) = ln|x|. To solve for y, take the tangent of both sides: y = tan(ln|x|).