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I want to ask you a question. For example I have an equation: $$u_{tt}-7u_{xx}-u_{x}=0 $$ To solve it I must first simplify it, right? I mean I must remove $u_x$. I suppose, that I must use next formulas: $$ u_{x} = e^{\lambda x + \mu t}(\lambda V + V_{x})$$ $$u_{xx} = e^{\lambda x + \mu t}(\lambda^{2} V + 2\lambda V_{x} + V_{xx})$$ $$u_{tt} = e^{\lambda x + \mu t}(\mu^{2} V + 2\mu V_{t} + V_{tt}) $$ Am I right? Will the primary conditions or\and boundary conditions change?

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When you substitute $e^{\lambda x+\mu t}V(x,t)$ for $u(x,t)$ in your PDE and get rid of the exponential, you find: $$V_{tt} -7\ V_{xx}-(14\lambda +1)\ V_x +2\mu\ V_t+(\mu^2-7\lambda^2-\lambda)\ V=0\; ;$$ if you want to get rid of $V_t$ and $V_x$ you have to choose $\lambda ,\mu$ s.t.: $$14\lambda +1=0\quad \text{and} \quad 2\mu=0\; ,$$ hence $\lambda=-1/14$ and $\mu=0$, so that: $$u(x,t)=e^{-x/14}\ V(x,t)$$ and your PDE becomes: $$V_{tt}-7\ V_{xx}+\frac{1}{28}\ V=0\; .$$

Now, it seems that you have the following IC/BC: $$\tag{IC} \begin{cases}u(x,0) = x - x^2 \\ u_t(x,0) = 0 \end{cases}$$ $$\tag{BC} \begin{cases}u(0,t) = 0 \\ u(1,t) = \sin(\pi t / 2)\end{cases}$$ and $t_{last} = 2$; such conditions in terms of $V$ read: $$\begin{cases} V(x,0) = e^{x/14}\ x\ (1 - x) & \text{, in } [0,1]\\ V_t(x,0) = 0 & \text{, in } [0,1]\end{cases} \qquad \text{and}\qquad \begin{cases} V(0,t) = 0 &\text{, in } [0,2]\\ V(1,t) = e^{1/14}\ \sin(\pi t / 2)&\text{, in } [0,2].\end{cases}$$

I have to say that, even if your substitution simplifies the PDE (for it cancels the term with the first derivative), it complicates the ICs; therefore that substitution doesn't seem useful.

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Yes yes yes. My answer is 98% same as yours) <br> math.stackexchange.com/questions/94292/… <br> But you give me acknowledgment, that my thoughts were right.... Fuf... It's good. Thank you very much. Because I was on 3-4 forums and 2 days no one has helped me. <br> P.S.: Is there any way to contact with you? –  ScienceSE Dec 27 '11 at 0:10
    
"PDE (for it cancels the term with the first derivative), it complicates the ICs" Professor said to do this. 4 days ago I tried did this without substitution and I had got very big and complicated koefficients etc. P.S.: I don't know how to complete solve this equation, I tried to do this last 3-4 days and I will trying to do this all the night today. Hopefully, that I'll the right answer for morning ) –  ScienceSE Dec 27 '11 at 0:25

You don't need those anzats, the equation is already amenable to a separation of variables via $u(t,x)=X(x)\cdot T(t)$. This substitution will give

$$ X(x)\cdot T''(t)- 7X''(x)\cdot T(t) - X'(x)\cdot T(t) = 0 $$ $$ X(x)\cdot T''(t)- (7X''(x) + X'(x))T(t)=0 $$ $$ \frac{T''(t)}{T(t)}=\frac{7X''(x) + X'(x)}{X(x)}=k,\hspace{5mm}k\in\mathbb{R} $$

Solve these single variable DE's with the help of your original BC/IC's.

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Thanks, I tried to do that, but I think it's not a good idea. Because I have inhomogeneous BC/IC's, without simplifying I will have a very complicated which I will not be able so solve ( –  ScienceSE Dec 25 '11 at 20:44
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@ScienceSE: If the problem lies in the specific BC/IC you are interested in, it seems rather odd that these are not even mentioned in your question. How to avoid making people able to answer one's question... –  Did Dec 26 '11 at 11:16
    
Sorry. I don't know english perfectly to describe all) I was in a consult-lesson today, the teacher said to do simplification first, so I'll try to do this. Actually sometimes I start to panic and nervous, when I don't know how to do somethink in mathematics, physics.. Because these things makes a great influence on my life... It would be nice if I knew a person, who knows this things good and have, for example, skype or something more.. –  ScienceSE Dec 26 '11 at 17:07

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