Initial-Value Problems - Differential Equations

So this is a separable differential equation, but it is also subject to an initial condition. This means that you have enough information so that there should not be a constant in the final answer.

You start off by getting all of the like terms on their respective sides, and then taking the anti-derivative. Your pre anti-derivative equation will look like:

Then taking the anti-derivative, you include a C value:

Then, using the initial condition given, we can solve for the value of C:

Solving for C, we get

which gives us a final answer of:

So this is a separable differential equation. We can think of as

and as .

Taking the anti-derivative once, we get:

Then using the initial condition

We get that

So Then taking the antiderivative one more time, we get:

and using the initial condition

we get the final answer of:

We have so that and . Solving for y,

and

which we can write because is just another arbitrary constant.

Plugging in our initial value, we have leaving us with a final answer of .

Note, this type of equation pops up frequently in the course and is potentially good to just memorize. For , we have

So this is a separable differential equation with a given initial value.

To start off, gather all of the like variables on separate sides.

Then integrate, and make sure to add a constant at the end

To solve for y, take the natural log, ln, of both sides

Be careful not to separate this, a log(a+b) can't be separated.

Plug in the initial condition to get:

So raising e to the power of both sides:

Solving for C:

giving us a final answer of:

So this is a separable differential equation with a given initial value.

To start off, gather all of the like variables on separate sides.

Notice that when you divide sec(y) to the other side, it will just be cos(y),

and the csc(x) on the bottom is equal to sin(x) on the top.

Integrating, we get:

so we can plug pi/4 into both x and y:

this gives us a C value of

In order to solve for y, we just need to take the arcsin of both sides:

So if you rearrange this equation, you will arrive at a separable differential equation by adding the to the other side:

Now, to solve this, multiply the dx to the other side and take the anti-derivative:

Then, after the anti-derivative, make sure to add the constant C:

Now, plug in the initial condition that y(0)=0, which will give you a C=0 as well. Then just take the square root, and you arrive at:

Then, to get the correct answer, simplify by factoring out an and pulling it outside of the square root to get:

So this is a separable differential equation with a given initial value.

To start off, gather all of the like variables on separate sides.

Then integrate, and make sure to add a constant at the end

Plug in the initial condition

Solving for C:

Which gives us:

Then taking the square root to solve for y, we get:

First identify what is known.

The general function is,

The initial value is six in mathematical terms is,

From here, substitute in the initial values into the function and solve for .

Finally, substitute the value found for into the original equation.

First identify what is known.

The general function is,

The initial value is six in mathematical terms is,

From here, substitute in the initial values into the function and solve for .

Finally, substitute the value found for into the original equation.

This type of problem will require an integrating factor. First, let's get it into a better form: . Once we have an equation in the form , we find (where the constant of integration is omitted because we only need one, arbitrary integrating factor).

Once we do this, we can see that

This is simply due to product rule, and then at the end, substitution of the original equation. Thus, as we know that , we can just integrate both sides to find y.

. A quick application of integration by parts with and tells us that the right hand side is . Dividing both sides by mu, we are left with . Plugging in the initial condition, we find giving us . Thus, we have a final answer of .