## Simple partial differential equation

Hello.

I have equation:

$$\frac{\partial T}{\partial t}-\frac{1}{2}\cdot \frac{(\partial)^2 T}{\partial x^2}=0$$

I calculated determinant:

$$\Delta=(-\frac{1}{2})^2)-4\cdot 1 \cdot 0 \Rightarrow \sqrt{\Delta}=\frac{1}{2} \\ (\frac{dT}{dt})_{1}=-\frac{1}{4} \\ (\frac{dT}{dt})_{2}=\frac{1}{4}$$

next

$$T=-\frac{1}{4}t+C_{1} \Rightarrow T+\frac{1}{4}t=C_{1} \\ T=\frac{1}{4}t+C_{2} \Rightarrow T-\frac{1}{4}t=C_{2}$$

I am add a new coefficients $$\eta$$ and $$\xi$$, then

$$\xi=\frac{1}{4}t+T\\ \eta=-\frac{1}{4}t+T$$

Then I calculated matrix jacobian's =$$\frac{1}{2}$$

Good?

I greet

Post edited

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 Quote by dexter90 Hello. I have equation: $$\frac{\partial T}{\partial t}-\frac{1}{2}\cdot \frac{(\partial)^2 T}{\partial x^2}=0$$ I calculated determinant: $$\Delta=(-\frac{1}{2})^2)-4\cdot 1 \cdot 0 \Rightarrow \sqrt{\Delta}=\sqrt{2} \\ (\frac{dT}{dt})_{1}=-\sqrt{2} \\ (\frac{dT}{dt})_{2}=\sqrt{2}$$
I'm not all that clear on what you are doing but that first statement is obviously untrue.
$$\Delta= (-\frac{1}{2})^2- 4\cdot 1 \cdot 0= \frac{1}{4}$$
so
$$\sqrt{\Delta}= \frac{1}{2}$$
not [itex]\sqrt{2}[itex]

 next $$T=-\sqrt{2}t+C_{1} \Rightarrow T+\sqrt{2}t=C_{1} \\ T=\sqrt{2}t+C_{2} \Rightarrow T-\sqrt{2}t=C_{2}$$ I am add a new coefficients $$\eta$$ and $$\xi$$, then $$\xi=\sqrt{2}t+T\\ \eta=-\sqrt{2}t+T$$ Then I calculated matrix jacobian's =$$2\sqrt{2}$$ Good? I greet
 Thanks, Of course, I made mistake in write. I would like solve partial differential equation but I dont have experience. I edited my post.