# Help with partial derivatives

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## Homework Statement

Hello I am given the equation:

ut - 2uxx = u
I was given other equations (boundary, eigenvalue equations) but i don't think I need that to solve this first part:

The book says to get rid of the zeroth order term by substituting u = exp(t)V(x,t). I tried to but I can't find a way to get rid of the zeroth order term! Any help would be appreciated.

## The Attempt at a Solution

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George Jones
Staff Emeritus
Gold Member
I tried to but ...

3. The Attempt at a Solution
Show your attempt at a solution.

lurflurf
Homework Helper
expand and simplify
##(\exp(t)V(x,t))_t-2(\exp(t)V(x,t))_{xx}=\exp(t)V(x,t)##

Ray Vickson
Homework Helper
Dearly Missed
expand and simplify
##(\exp(t)V(x,t))_t-2(\exp(t)V(x,t))_{xx}=\exp(t)V(x,t)##
So, now expand the derivatives using the product rule, etc.

Sorry for the late reply. I have the following after expanding using the product rule:

(1) etV(x,t) + etV(x,t)t -2etV(x,t)xx - etV(x,t) = 0

I can simplify to:

(2) etV(x,t)t -2etV(x,t)xx = 0

(3) Now I will let V(x,t) = X(x)T(t)

Plugging in to (2) I get:

(4) etX(x)T'(t) - 2etX''(x)T(t) = 0

I can now separate the variables but they will both end up with e^t. Is there a way to get rid of the e^t or am I doing something wrong?[/SUP][/SUP]

Ray Vickson
Homework Helper
Dearly Missed
Sorry for the late reply. I have the following after expanding using the product rule:

(1) etV(x,t) + etV(x,t)t -2etV(x,t)xx - etV(x,t) = 0

I can simplify to:

(2) etV(x,t)t -2etV(x,t)xx = 0

(3) Now I will let V(x,t) = X(x)T(t)

Plugging in to (2) I get:

(4) etX(x)T'(t) - 2etX''(x)T(t) = 0

I can now separate the variables but they will both end up with e^t. Is there a way to get rid of the e^t or am I doing something wrong?[/SUP][/SUP]
What is stopping you from dividing equation (4) by the nonzero quantity ##e^t?##

BTW: you will save yourself hours of complex, frustrating and and error-prone typing, just by using LaTeX to typeset your expressions and equations. For example, (4) is a snap to write and looks better as well:
$$e^t X(x) T'(t) - 2 e^t X''(x) T(t)=0 \hspace{4ex}(4)$$
Just right-click on the image and ask for a display of math as tex commands.

jim mcnamara
What is stopping you from dividing equation (4) by the nonzero quantity ##e^t?##

BTW: you will save yourself hours of complex, frustrating and and error-prone typing, just by using LaTeX to typeset your expressions and equations. For example, (4) is a snap to write and looks better as well:
$$e^t X(x) T'(t) - 2 e^t X''(x) T(t)=0 \hspace{4ex}(4)$$
Just right-click on the image and ask for a display of math as tex commands.
I am new to mathematics via forums. Thank you for twlltelme about latex. As for the problem, I realized that I could divide the e^t (I hadn't slept for a while). Thank you for the answers everyone! On to the next challenge!