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

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