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Partial differential equations problem - finding the general solution

Another

Member
Feb 11, 2018
40
\(\displaystyle 4\frac{\partial u}{\partial t}+\frac{\partial u}{\partial x} = 3u\) , \(\displaystyle u(x,0)=4e^{-x}-e^{-5x}\)

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
\(\displaystyle \frac{\partial X}{X \partial x} =\left( 3 - 4\frac{\partial T}{T \partial t} \right) = k \)
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\(\displaystyle \frac{\partial X}{X \partial x} = k \)
\(\displaystyle \frac{d X}{X} = k dx\)
\(\displaystyle X = C_1e^{kx}\)
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\(\displaystyle \left( 3 - 4\frac{\partial T}{T \partial t} \right) = k \)
\(\displaystyle 4\frac{\partial T}{T \partial t}= 3 - k \)
\(\displaystyle \frac{\partial T}{T \partial t}= \frac{1}{4}(3 - k) \)
\(\displaystyle \frac{d T}{T}= \frac{1}{4}(3 - k) dt \)
\(\displaystyle T = C_2e^{\frac{1}{4}(3 - k) t} \)
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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
Feb 7, 2012
2,770
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
Feb 11, 2018
40
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