# Another partial differential equation problem

## Homework Statement

Derived the general solution of the given equation by using an appropriate change of variables
5.) du/dt-2(du/dx)=2

## Homework Equations

du/dx=du/d(alpha)*d(alpha)/dx+du/d(beta)*d(beta)/dx
du/dt= du/d(alpha)*d(alpha)/dt+du/d(beta)*d(beta)/dt

## The Attempt at a Solution

not really sure how to determine du/d(alpha),d(alpha)/dx,d(beta)/dx ,d(alpha)/dt,d(beta)/dt

book says alpha=x+2t, beta=x

## Homework Statement

Derived the general solution of the given equation by using an appropriate change of variables
5.) du/dt-2(du/dx)=2

## Homework Equations

du/dx=du/d(alpha)*d(alpha)/dx+du/d(beta)*d(beta)/dx
du/dt= du/d(alpha)*d(alpha)/dt+du/d(beta)*d(beta)/dt

## The Attempt at a Solution

not really sure how to determine du/d(alpha),d(alpha)/dx,d(beta)/dx ,d(alpha)/dt,d(beta)/dt

book says alpha=x+2t, beta=x
$$\frac{\partial\alpha}{\partial x}=1,\frac{\partial\beta}{\partial x}=1 so $\frac{\partial u}{\partial x}=\frac{\partial u}{\partial\alpha}\frac{\partial\alpha}{\partial x}+\frac{\partial u}{\partial\beta}\frac{\partial\beta} {\partial x}=\frac{\partialu}{\partialx}=\frac{\partial u}{\partial\alpha}+\frac{\partial u}{\partial\beta}$$$
did that help?

$$\frac{\partial\alpha}{\partial x}=1,\frac{\partial\beta}{\partial x}=1 so $\frac{\partial u}{\partial x}=\frac{\partial u}{\partial\alpha}\frac{\partial\alpha}{\partial x}+\frac{\partial u}{\partial\beta}\frac{\partial\beta} {\partial x}=\frac{\partialu}{\partialx}=\frac{\partial u}{\partial\alpha}+\frac{\partial u}{\partial\beta}$$$
did that help?

I probably should have said in the back of the book, they give you alpha and beta. They don't give you the equations for alpha and beta in the problem. I don't understand how to determine alpha and beta.