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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.

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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