1. ## Differential Equation

Hello everyone!

Can somebody help me out with this one please...

2. It's a "Show That". All you need is a couple of derivatives and some algebra.

Let's see what you get for the 1st and 2nd derivative of the provided expression.

3. for part a) Find $\frac{dy}{dx}$ and $\frac{d^2y}{dx^2}$ and substitute it into the equation given you should find that the LHS=RHS which shows its a solution.

for the second part you need to solve the complimentary function
$\frac{d^2y}{dx^2}-2\frac{dy}{dx}+y=0$

this would have an auxilary quadratic of
$\lambda^2-2\lambda+1=0$

solving that you get $\lambda=1$ as a repeated root so the solution to the complimentary function is

$y=e^x(A+Bx)$ where A and B are the constants of integration.

Putting this together with the solution we were given in part a) we have

$y=e^x(A+Bx)+\frac{1}{2}x^2e^x$

so all we have to do is find the constants from the given boundary conditions.

I get A=1 and B=1 do post again if you have any problem with doing that or understanding what I've written

Hope that helps

Simon

4. I couldnt get the ansa as e^x

5. your differentiating wrong do you know the product rule?

When u and v are functions of x
$
\frac{d}{dx}(uv)=u\frac{dv}{dx}+v\frac{du}{dx}$

in this case we would have $u=\frac{1}{2}x^2$ and $v=e^x$

so using the product rule

$\frac{dy}{dx}=\frac{1}{2}x^2e^x+xe^x$

can you do it the second time to get $\frac{d^2y}{dx^2}$ now?

6. yeh, I get really confused knowing when to use which rule I can see why its the product rule here but would I now use the chain rule to diffeentiate again because the result isnt a product now

7. no you would use the product rule but twice remember that

$\frac{d}{dx}(\frac{1}{2}x^2e^x+xe^x)=\frac{d}{dx}( \frac{1}{2}x^2e^x)+\frac{d}{dx}(xe^x)$

8. I think i went wrong sumplace...

9. Originally Posted by Sweeties
I think i went wrong sumplace...
Hello,

In the last line :

$xe^x+xe^x+e^x=(2x+1)e^x \neq 3xe^x$

Also, be careful when using tex for multiplications, don't use x because it's just like the variable and we can easily get confused. You can use $\cdot$ ( \cdot) or $\times$ ( \times)

Plus, you made some typos like d/dx (uv)=(xe^x), which is false to write. I guess you wanted to say :

$\frac{d}{dx} (xe^x)=\frac{d}{dx} (uv)$

Want me to try showing you how to write all this stuff, or is it ok ?

10. Thanks Moo
I am using word 2007 with microsoft math add-in, just getting used to it
i havent got latex on this computer, been meaning to download it!

I think I got it now.

11. Originally Posted by Sweeties
Thanks Moo
I am using word 2007 with microsoft math add-in, just getting used to it
i havent got latex on this computer, been meaning to download it!

I think I got it now.
You just have to use  here

And know some things about the latex code (there are topics in the latex help forum )

12. Originally Posted by Moo
You just have to use  here

And know some things about the latex code (there are topics in the latex help forum )
Wicked!!! I just went through the tutorial. Thanks moo!!!