Question

dB/dx=B/t2
t is constant.if we solve this we get
B=B0 ex/t+B0e-x/t
if the graph is DECAYING EXPONENTIALY then can we ignore the B0 ex/t term?

Answer 1

Answer on Question #62809 – Math – Differential Equations

Question

$\mathrm{dB/dx = B/t2}$

t is constant. if we solve this we get $\mathrm{B} = \mathrm{B}0\mathrm{ex/t} + \mathrm{B}0\mathrm{e - x/t}$ if the graph is DECAYING EXPONENTIALLY then can we ignore the B0 ex/t term?

Solution

$B = B_{0}e^{x / t} + B_{0}e^{-x / t}$ is not a solution of equation

$$\frac {d B}{d x} = \frac {B}{t ^ {2}}$$

Because

$$\frac {d}{d x} \Big (B _ {0} e ^ {\frac {x}{t}} + B _ {0} e ^ {- \frac {x}{t}} \Big) = B _ {0} \left(\frac {1}{t} e ^ {\frac {x}{t}} - \frac {1}{t} e ^ {- \frac {x}{t}}\right) = B _ {0} \frac {1}{t} \Big (e ^ {\frac {x}{t}} - e ^ {- \frac {x}{t}} \Big) \neq \frac {B}{t ^ {2}} = \frac {(B _ {0} e ^ {x / t} + B _ {0} e ^ {- x / t})}{t ^ {2}}$$

But, $B = B_0e^{x / t} + B_0e^{-x / t}$ is the solution of the equation

$$\frac {d ^ {2} B}{d x ^ {2}} = \frac {B}{t ^ {2}}$$

Because

$$\frac {d ^ {2} B}{d x ^ {2}} = \frac {d}{d x} \left(B _ {0} \frac {1}{t} \left(e ^ {\frac {x}{t}} - e ^ {- \frac {x}{t}}\right)\right) = B _ {0} \frac {1}{t} \frac {d}{d x} \left(e ^ {\frac {x}{t}} - e ^ {- \frac {x}{t}}\right) = B _ {0} \frac {1}{t} \left(\frac {1}{t} e ^ {\frac {x}{t}} + \frac {1}{t} e ^ {- \frac {x}{t}}\right) =$$ $$= B _ {0} \frac {1}{t ^ {2}} \left(e ^ {\frac {x}{t}} + e ^ {- \frac {x}{t}}\right) = \frac {B}{t ^ {2}}$$

Let's solve equation

$$\frac {d ^ {2} B}{d x ^ {2}} = \frac {B}{t ^ {2}}$$

We will find the solution in the form

$$B = e ^ {\lambda x}$$

Substituting into the equation yields

$$\frac {d ^ {2} B}{d x ^ {2}} = \lambda^ {2} e ^ {\lambda x} = \frac {1}{t ^ {2}} e ^ {\lambda x}$$

From this we obtain the characteristic equation: $\lambda^2 = \frac{1}{t^2}$. The solutions of this equation are

$$\lambda = \pm \frac {1}{t}.$$

Thus we have two particular solutions $B_{1} = e^{x / t}$ and $B_{2} = e^{-x / t}$.

The general solution of the original equation is the linear combination of the particular solutions:

$$B = C e ^ {x / t} + D e ^ {- x / t}$$

where $C$ and $D$ are some constants which we chose depending on initial or boundaries conditions, if $B \to 0$ when $x \to \infty$, we have to take $C = 0$, so we have $B = D e^{-x / t}$ or $B = B_0 e^{-x / t}$ where $B = B_0$ at $x = 0$

**Answer**: $B = B_{0}e^{-x / t}$.

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