# Partial differential equations problem - finding the general solution

#### Another

##### Member
$$\displaystyle 4\frac{\partial u}{\partial t}+\frac{\partial u}{\partial x} = 3u$$ , $$\displaystyle u(x,0)=4e^{-x}-e^{-5x}$$

let $$\displaystyle U =X(x)T(t)$$

so

$$\displaystyle 4X\frac{\partial T}{\partial t}+T\frac{\partial X}{\partial x} = 3XT$$
$$\displaystyle 4\frac{\partial T}{T \partial t}+\frac{\partial X}{X \partial x} = 3$$
$$\displaystyle \left( 4\frac{\partial T}{T \partial t}-3 \right) +\frac{\partial X}{X \partial x} = 0$$

let K = constant
$$\displaystyle \frac{\partial X}{X \partial x} =\left( 3 - 4\frac{\partial T}{T \partial t} \right) = k$$
_________________________________________________________________________________

$$\displaystyle \frac{\partial X}{X \partial x} = k$$
$$\displaystyle \frac{d X}{X} = k dx$$
$$\displaystyle X = C_1e^{kx}$$
_________________________________________________________________________________

$$\displaystyle \left( 3 - 4\frac{\partial T}{T \partial t} \right) = k$$
$$\displaystyle 4\frac{\partial T}{T \partial t}= 3 - k$$
$$\displaystyle \frac{\partial T}{T \partial t}= \frac{1}{4}(3 - k)$$
$$\displaystyle \frac{d T}{T}= \frac{1}{4}(3 - k) dt$$
$$\displaystyle T = C_2e^{\frac{1}{4}(3 - k) t}$$
_________________________________________________________________________________

general solution
$$\displaystyle u(x,t) = C e^{kx}e^{\frac{1}{4}(3 - k) t}$$ then $$\displaystyle C=C_1C_2$$

$$\displaystyle u(x,0) = C e^{kx} = 4e^{-x}-e^{-5x}$$ <<How do I solve this equation?

#### Opalg

##### MHB Oldtimer
Staff member
general solution
$$\displaystyle u(x,t) = C e^{kx}e^{\frac{1}{4}(3 - k) t}$$ then $$\displaystyle C=C_1C_2$$

$$\displaystyle u(x,0) = C e^{kx} = 4e^{-x}-e^{-5x}$$ <<How do I solve this equation?
You have shown that $$\displaystyle u(x,t) = C e^{kx}e^{\frac{1}{4}(3 - k) t}$$ is a solution. But it is not the general solution. Instead, it provides a solution for every value of $k$. So a more general solution would be given by a sum of those solutions, for different values of $k$: $u(x,t) = \sum_k C_k e^{kx}e^{\frac{1}{4}(3 - k) t}.$ In particular, you could take the solutions for $k=-1$ and $k=-5$, and use their sum as a solution.

#### Another

##### Member
You have shown that $$\displaystyle u(x,t) = C e^{kx}e^{\frac{1}{4}(3 - k) t}$$ is a solution. But it is not the general solution. Instead, it provides a solution for every value of $k$. So a more general solution would be given by a sum of those solutions, for different values of $k$: $u(x,t) = \sum_k C_k e^{kx}e^{\frac{1}{4}(3 - k) t}.$ In particular, you could take the solutions for $k=-1$ and $k=-5$, and use their sum as a solution.
Great! thank you