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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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    "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 )2011-12-27
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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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    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..2011-12-26