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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?