Working Out Answer Checks - DE

[erok81]

Homework Statement

The symbolic solution of y'=-2xy, y(0)=2 is y(x)=2e^{-x^2}. Display the details of the linear integrating factor method derivation of this symbolic solution, plus a full answer check.

Homework Equations

Linear integration factor method uses the standard for y'+p(x)y=q(x)

The Attempt at a Solution

I can solve this no problem. I am having trouble doing the answer checks on these. With previous sections you could just take the derivative of both sides and end up with the original statement. However with this, I cannot seem to make it anywhere close to the original problem.

This for example taking the solution from above:
y(x)=2e^{-x^2}

If I take the derivative of both sides I end up with:

y'=-4e^{-x^2}

Which isn't even remotely close.

What concept am I missing here?

 

[LCKurtz]

You are forgetting the chain rule when you differentiate.

 

[erok81]

Hmm...I thought I did.

Starting with:

y=2e^{-x^{2}}

Without simplifying I get y'=2e^{-x^{2}}(-2x)

Am I really doing the derivative of something that simple wrong?

I threw it into wolframalpha.com and maple and got the same answers.

 

[LCKurtz]

Hmm...I thought I did.

Starting with:

y=2e^{-x^{2}}

Without simplifying I get y'=2e^{-x^{2}}(-2x)

Am I really doing the derivative of something that simple wrong?

I threw it into wolframalpha.com and maple and got the same answers.

No, that is right but it isn't what you had in your earlier post. Anyway, all you have left to do is to verify that that value of y' equals -2xy and that y satisfies the BC's. Pretty easy now.

 

[erok81]

Oh yeah, I left off the extra x when I typed that out.

I guess the part I am confused with is how I go from something with an exponential and no y to no exponential and a y.

This is the first class I've had to do answer checks so I've never learned exactly how to do them.

 

[LCKurtz]

Oh yeah, I left off the extra x when I typed that out.

I guess the part I am confused with is how I go from something with an exponential and no y to no exponential and a y.

This is the first class I've had to do answer checks so I've never learned exactly how to do them.

You have the DE y' = -2xy and you are trying to show y=2e^{-x^2} is a solution. You have calculated y'=-4xe^{-x^2}. All you have to do is plug your expressions for y and y' into the DE and see it reduces it to an identity, which means the equation is satisfied.

 

[erok81]

So I tried this...

Here is what I have.

DE = y'=-2xy
y=2e^{-x^{2}}
y'=-4xe^{-x^{2}}

Taking the original DE and subbing in what I think is correct.

-4xe^{-x^{2}}=-2x(2e^{-x^{2}})

Then.
-4xe^{-x^{2}}=-4xe^{-x^{2}}

So both values are equal but I don't see an identity.

 

[LCKurtz]

Here's a definition for you:

Identity Equation: An equation which is true for every value of the variable is called an identity equation.

Now do you see one?

 

[erok81]

Ooooh, yes I get it now!

I was thinking something like trig identity.

Perfect. Thanks for the help. I am glad this class requires these on a lot of the problems. One tends to learn A LOT more when they do the answer one way then back again.