# Calculus 1 : How to find solutions to differential equations

## Example Questions

### Example Question #51 : Differential Equations

Find the general solution for the following differential equation:

Explanation:

First we must rearrange this separable differential equation so that we can place    alone on one side and    on the other side with the terms involving    and any constants. We then integrate each side with respect to the appropriate variable and solve the result for    to find the general solution of the differential equation:

Remember when integrating, we increase the exponent by one and then divide the whole term by the value of the new exponent. Will we need to integrate each term that contains  in this fashion.

### Example Question #52 : Differential Equations

Find the general solution for the following differential equation:

Explanation:

First we must multiply    to the right side of the equation so we have the    terms with    and the    terms with  .  We can then integrate each side with respect to the appropriate variable and then solve the result for    to find the general solution for the differential equation:

Remember when integrating, we increase the exponent by one and then divide the term by that new exponent value. We will do this with each term on both sides.

### Example Question #53 : Differential Equations

Determine the general solution to the following differential equation:

Explanation:

In order to separate the differential equation such that    is with the    terms and    is with the    terms, we must divide both sides of the equation by    and then multiply both sides by  .  We can then integrate each side with respect to the appropriate variable, and then solve for    to find the general solution to the differential equation:

### Example Question #54 : Differential Equations

Determine the general solution to the following differential equation:

Explanation:

For this separable differential equation, we can see that if we cross multiply we will have the    term with  ,  and the    term with  .  After getting the equation into this form, we can integrate each side with respect to the appropriate variable, and then solve for    to find the general solution to the differential equation:

Remember when integrating, we increase the exponent by one and then divide the term by that new exponent value. We will do this with each term on both sides.

### Example Question #55 : Differential Equations

Find the particular solution for the following initial value problem:

Explanation:

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:

Remember when integrating, we can use the power rule. This means to increase the exponent by one and then divide the term by that new exponent value. We will do this with each term on both sides.

Once we have found the general solution we plug in the initial conditions to solve for C.

### Example Question #56 : Differential Equations

Find the particular solution for the following differential equation:

Explanation:

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. For this problem we can see that the desired arrangement is achieved by simply cross multiplying the differential equation. 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:

Remember when integrating, we can use the power rule. The power rule states to increase the exponent by one and then divide the term by that new exponent value. We will do this with each term on both sides.

Now that we have the general solution we can plug in the initial values and solve for C.

### Example Question #51 : Differential Equations

Find the first derivative of the following:

Explanation:

This problem requires the use of the product rule.

The product rule tells us that the derivative of,

is .

In this problem,  is , and  is .

Therefore,

, and  = , so

, and factoring out  gives us .

### Example Question #51 : Differential Equations

Determine the general solution of  , if possible.

Explanation:

To find the general solution of , rewrite this in the form of its characteristic equation. This will be a quadratic.

Factorize and solve for the roots.

Since the roots are distinct, write the specific form of the general solution stated by one of the three homogenous equation rules.

Substitute the roots in the formula to obtain the general solution.

### Example Question #51 : Differential Equations

Determine the general solution of , if possible.

Explanation:

To solve for the general solution of , convert this to its characteristic equation and solve for the roots.

Write the specific form of the general solution stated by one of the three homogenous equation rules for distinct roots.

Substitute the roots to obtain the general solution.

### Example Question #44 : Solutions To Differential Equations

Let . Find .

(Hint: Use differential equations.)

Explanation:

Let . Find .

(Hint: Use differential equations.)