### All Calculus 1 Resources

## Example Questions

### Example Question #1332 : Functions

Find the equilibrium values for the following differential equation:

**Possible Answers:**

**Correct answer:**

To find the equilibrium values, substitute and solve for . The equilibrium values are the solutions of the differential equation in constant form.

### Example Question #42 : Differential Equations

Suppose is greater than zero. Solve the differential equation:

**Possible Answers:**

**Correct answer:**

Rewrite so that the same variables are aligned correctly on the left and right of the equal sign.

Integrate both sides and solve for .

### Example Question #31 : How To Find Solutions To Differential Equations

Find the function whose slope at the point is and passes through the point .

**Possible Answers:**

**Correct answer:**

This question requires us to use differential equations. Begin as follows:

Now, we know that y must pass through the point (4,5), so we can use this point to find c

So our function is as follows:

### Example Question #43 : Differential Equations

Solve:

**Possible Answers:**

**Correct answer:**

Multiply and divide on both sides of the equation .

Integrate both sides.

Use base to eliminate the natural log.

### Example Question #31 : How To Find Solutions To Differential Equations

.

Calculate

**Possible Answers:**

**Correct answer:**

Remember that the derivative of .

.

Now plug in the to find the corresponding values.

Substitute these into the desired formula.

### Example Question #32 : How To Find Solutions To Differential Equations

Determine the general solution to the following differential equation:

**Possible Answers:**

**Correct answer:**

This is a separable differential equation, which means we can separate and , placing each on one side of the equation with its corresponding terms. There are no terms containing , so we simply place alone on one side of the equation and on the other side of the equation with any terms. We can then integrate each side with respect to the appropriate variable, which gives us an equation for that is the general solution to the differential equation:

### Example Question #47 : Differential Equations

Consider the chemical reaction .

Initially there are moles of . The change in concentration of occurs at the following rate.

Where .

Find at minutes.

**Possible Answers:**

**Correct answer:**

Rearrange the equation to have all the terms on one side and the rest of the terms on the other side.

From here we need to take the integral of the function.

Where C is some constant.

We know that the initial concentration of A is 10 mols. In otherwords, at

Solving for gives

.

Plugging everything in gives

.

### Example Question #48 : Differential Equations

Find the particular solution for the following initial value problem:

**Possible Answers:**

**Correct answer:**

To find the particular solution, we start by finding the general solution. First we rearrange the differential equation such that is on one side with any terms and is on the other side with any terms. We can then integrate each side with respect to the appropriate variable and solve for to find the general solution for the differential equation. Finally, we plug in the given initial condition to determine the value of the constant, which gives us the particular solution:

### Example Question #49 : Differential Equations

Find the particular solution for the following differential equation:

**Possible Answers:**

**Correct answer:**

To find the particular solution, we start by finding the general solution. First we rearrange the differential equation such that is on one side with any terms and is on the other side with any terms. We can then integrate each side with respect to the appropriate variable and solve for to find the general solution for the differential equation. Finally, we plug in the and values of the given point to determine the value of the constant, which gives us the particular solution:

### Example Question #50 : Differential Equations

Determine the general solution to the following differential equation:

**Possible Answers:**

**Correct answer:**

This is a separable differential equation, which means we are able to separate and , placing them on opposite sides of the equation with their corresponding variables. There are no terms, so we place alone on one side, and on the other side with the terms containing . We can then integrate each side with respect to the appropriate variable, which gives an equation for that is the general solution to the differential equation:

Remember when we integrate, we increase the exponent by one and then divide the term by the value of the new exponent. We will need to integrate each term that contains a .