Hello everyone!
Can somebody help me out with this one please...
for part a) Find $\displaystyle \frac{dy}{dx}$ and $\displaystyle \frac{d^2y}{dx^2}$ 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
$\displaystyle \frac{d^2y}{dx^2}-2\frac{dy}{dx}+y=0$
this would have an auxilary quadratic of
$\displaystyle \lambda^2-2\lambda+1=0$
solving that you get $\displaystyle \lambda=1$ as a repeated root so the solution to the complimentary function is
$\displaystyle y=e^x(A+Bx)$ where A and B are the constants of integration.
Putting this together with the solution we were given in part a) we have
$\displaystyle y=e^x(A+Bx)+\frac{1}{2}x^2e^x$
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
your differentiating wrong do you know the product rule?
When u and v are functions of x
$\displaystyle
\frac{d}{dx}(uv)=u\frac{dv}{dx}+v\frac{du}{dx}$
in this case we would have $\displaystyle u=\frac{1}{2}x^2$ and $\displaystyle v=e^x$
so using the product rule
$\displaystyle \frac{dy}{dx}=\frac{1}{2}x^2e^x+xe^x$
can you do it the second time to get $\displaystyle \frac{d^2y}{dx^2}$ now?
Hello,
In the last line :
$\displaystyle xe^x+xe^x+e^x=(2x+1)e^x \neq 3xe^x$
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 $\displaystyle \cdot$ ( \cdot) or $\displaystyle \times$ ( \times)
Plus, you made some typos like d/dx (uv)=(xe^x), which is false to write. I guess you wanted to say :
$\displaystyle \frac{d}{dx} (xe^x)=\frac{d}{dx} (uv)$
Want me to try showing you how to write all this stuff, or is it ok ?