Could anyone help?

[linkj6]

I have no idea what to do in this subject. My professor hasn't really explained anything to us and the book is less than helpful. A question I need help on says:

"Determine by inspection at least one solution of the given differential equation. That is, use your knowledge of derivatives to make an intelligent guess. Then test your hypothesis."
Then it gives me this equation: (y')^2 + y^2 = 1

I have no clue what this is called or where to begin. Even if I did know where to start I'm not really sure what I'm looking for (a solution but that's as vague as it gets). Also, if anyone knows of some really good references that explain how to do anything and everything with differential equations please let me know. Any help would be greatly appreciated.

 

[mmm4444bot]

(y')^2 + y^2 = 1

I have no clue what this is called or where to begin. Even if I did know where to start I'm not really sure what I'm looking for.

That's called a differential equation because it contains a derivative.

You're familiar with functions, yes? The given equation contains two functions, and they are added together.

y` is one function.

y is another function.

It's no different than writing: f`(x)^2 + f(x)^2 = 1.

Yet, it's standard to use the name y instead of f, when writing differential equations.

Also, the notion that y is a function of x is implied. In other words, the symbol y means y(x), and the symbol y` means y`(x). We leave off the independent variable to simplify the notation, and just write y and y`.

So, you are looking for a function y(x) = some expression.

But, to be a solution of this differential equation, function y must lead to a sum of 1 when its square is added to the square of its first derivative.

y^2 + (y`)^2 = 1

Here are some hints:

What is the first derivative of any constant function?

Does that answer give you any clues about what the value of y^2 must be, in the differential equation, in order for the sum to be 1?

Here's a BIG hint for a different approach.

In trigonometry, there is a well-known identity that involves a sum of squares that always equals 1. And the sine and cosine functions just happen to be derivatives of each other, if we ignore sign changes. But everything's positive when squared, right? So, the signs don't matter.

The given instruction stipulates "Determine by inspection". In mathematics, that phrase generally means doing it in your head, or somehow recognizing the solution with little to no paperwork, based on prior knowledge.

If it's true that you really don't know about any of this stuff (i.e., the prerequisite material required for understanding these concepts), then there is little chance that you'll be able to do anything by inspection.

 

[linkj6]

I think I understand this one. There are two answers correct? sine and cosine?

 

[mmm4444bot]

There are two answers correct? sine and cosine?

There are many possible answers, in this exercise.

You can write as many of these answers as you like, but you're only required to write one answer because they stated "at least one solution".

By "inspection", I see two constant functions y(x) that are solutions to the given differential equation.

I also see an infinite number of trigonometric functions y(x) that each involve either the sine function or the cosine function. (There are infinite solutions of this type because sine and cosine are periodic. In other words, adding an arbitrary multiple of Pi to the independent variable doesn't alter these functions' output.)

Your answer needs to be in the form of a function definition: y(x) = some expression.

Simply writing either of the nouns "sine" or "cosine" is not good enough.

ALSO, you're required to "test your hypothesis".

In other words, after you decide on a function definition for y(x), you need to show that it solves the differential equation.

 

[linkj6]

Ok. I think I understand. Thanks for the help.

I have another question. It looks like a different type of problem and I'm a little lost. It's:

Verify that y(x) = e^(x^2) * integral from 0 to x (e^(-t^2)dt)
is a solution to dy/dx = 2xy + 1

I'm not sure how to go about doing this. Any help would be appreciated.
Also, sorry if I shouldn't be posting two questions in the same thread. I figured I save one extra topic from being posted.

 

[mmm4444bot]

sorry if I shouldn't be posting two questions in the same thread. I figured I save one extra topic from being posted.

I'm confident that nobody really cares how many new topics are posted at this site.

I'm not able to help you with your second exercise.

Other people may not see it "buried" at the end of this thread, so my suggestion is to repost it by starting a new topic.

Also, here's one possible solution to your first exercise.

y(x) = 1

 

[Deleted member 4993]

Ok. I think I understand. Thanks for the help.

I have another question. It looks like a different type of problem and I'm a little lost. It's:

Verify that y(x) = e^(x^2) * integral from 0 to x (e^(-t^2)dt) <<< Start with finding [sub:iio2l3yq]\(\displaystyle \frac{dy}{dx}\)[/sub:iio2l3yq]
is a solution to dy/dx = 2xy + 1

I'm not sure how to go about doing this. Any help would be appreciated.
Also, sorry if I shouldn't be posting two questions in the same thread. I figured I save one extra topic from being posted.