Question

Find the solution to the differential equation dz/dt = 7te^4z that passes through the origin.

Answer 1

Answer:

$z=(\frac{-1}{4} )ln|1-14t^{2}|$

Step-by-step explanation:

from the equation, $\frac{dz}{dt}=7te^{4z}$.

we can give different approach to the equation, but to make it simple and direct, let separate the equation by bringing all like terms to the same side i.e

$\frac{dz}{e^{4z}}=7tdt\\e^{-4z} dz=7tdt$.

if we integrate both side,

$\int\limits^a_b{e^{-4z} } \,dz =\int\limits^a_b {7t} \,dt$

$-1/4e^{-4z} +c_{1}=7/2t^{2} +c_{2}\\-1/4e^{-4z}= 7/2t^{2} +c_{2}-c_{1}\\let c_{2}-c_{1}=c \\-1/4e^{-4z}= 7/2t^{2} +c$

since the equation passes through the origin, and at the origin z=0 and t=0

we substitute this values and solve for the constant c

$e^{-4*0}= 7/2*0 +c\\c=1$.

If we substitute the value of c into the equation we arrive at

$(-1/4)e^{-4z}= (7/2)t^{2}+1\\ e^{-4z}=1-14t^{2}$

if we the the natural logarithm of both sides, we arrive at

$-4z=ln|1-14t^{2}|\\z=(\frac{-1}{4} )ln|1-14t^{2}|$

Answer 2

Using separation of variables, it is found that the solution to the differential equation is:

• $z(t) = -\frac{\ln{-14t^2 + 1}}{4}$

The differential equation is:

$\frac{dz}{dt} = 7te^{4z}$

How is separation of variables applied?

• We have two variables, z and t, hence, everything with z goes to one side, everything with t to the other side, and both sides are integrated.

Then:

$\frac{dz}{e^{4z}} = 7t dt$

$e^{-4z} dz = 7t dt$

$\int e^{-4z} dz = \int 7t dt$

$-\frac{e^{-4z}}{4} = \frac{7t^2}{2} + K$

$e^{-4z} = -14t^2 + K$

$\ln{e^{-4z}} = \ln{-14t^2 + K}$

$-4z = \ln{-14t^2 + K}$

$z = -\frac{\ln{-14t^2 + K}}{4}$

It goes through the origin, hence when z = 0, t = 0 and this is used to find K.

$z = -\frac{\ln{-14t^2 + K}}{4}$

$0 = -\frac{\ln{-14(0)^2 + K}}{4}$

$\ln{K} = 0$

$e^{\ln{K}} = e^0$

$K = 1$

Hence, the solution is:

$z(t) = -\frac{\ln{-14t^2 + 1}}{4}$

To learn more about separation of variables, you can check https://brainly.com/question/14318343