MHB Partial differential equations problem - finding the general solution

[Another1]

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

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

so

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

let K = constant
$$\frac{\partial X}{X \partial x} =\left( 3 - 4\frac{\partial T}{T \partial t} \right) = k $$
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$$\frac{\partial X}{X \partial x} = k $$
$$\frac{d X}{X} = k dx$$
$$ X = C_1e^{kx}$$
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$$\left( 3 - 4\frac{\partial T}{T \partial t} \right) = k $$
$$ 4\frac{\partial T}{T \partial t}= 3 - k $$
$$ \frac{\partial T}{T \partial t}= \frac{1}{4}(3 - k) $$
$$ \frac{d T}{T}= \frac{1}{4}(3 - k) dt $$
$$ T = C_2e^{\frac{1}{4}(3 - k) t} $$
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general solution
$$ u(x,t) = C e^{kx}e^{\frac{1}{4}(3 - k) t} $$ then $$ C=C_1C_2 $$

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

 

[Opalg]

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

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

You have shown that $$ u(x,t) = C e^{kx}e^{\frac{1}{4}(3 - k) t} $$ 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.

 

[Another1]

You have shown that $$ u(x,t) = C e^{kx}e^{\frac{1}{4}(3 - k) t} $$ 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