Respuesta :
Answer:
[tex]C_{A}=\frac{e^{\sqrt{\frac{10}{3} }z}}{10}[/tex]
Step-by-step explanation:
First you need to rearrange the original equation in order to have an equation with a general form:
[tex]D \frac{d^{2}C_{A}}{d^{2}z} -kC_{A}=0[/tex]
We divide everything by D, so we have:
[tex]\frac{d^{2}C_{A}}{d^{2}z} -\frac{k}{D} C_{A}=0[/tex]
We know that k and D are constants, so we can leave them as just one constant that gives as result:
[tex]\frac{10}{3} =\frac{k}{D}[/tex]
Then we proceed to change terms in the equation son it looks friendlier:
[tex]y''-\frac{10}{3}y=0[/tex]
Knowing that [tex]y=C_{A}[/tex]
Then we give the general solution for the differential equation of second order:
[tex]y=C_{1}e^{\sqrt{\frac{10}{3} } z} + C_{2}e^{-\sqrt{\frac{10}{3} } z}[/tex]
The following is knowing the frontier values so we can calculate the constants of the equation:
[tex]z=0; y=0.1\\z=z; y=0[/tex]
With these values, we can replace in the general solution to find the final equation:
[tex]0.1=C_{1}e^{\sqrt{\frac{10}{3} }0} + C_{2}e^{-\sqrt{\frac{10}{3} }0}\\0.1=C_{1} + C_{2}[/tex]
And with the other values:
[tex]0=C_{1}e^{\sqrt{\frac{10}{3} }z} + C_{2}e^{-\sqrt{\frac{10}{3} }z}[/tex]
Then we solve the equation system, as we have two unknowns, the constants, and two equations:
[tex]C_{1}=0.1-C_{2}\\0=(0.1-C_{2})e^{\sqrt{\frac{10}{3} }z} +C_{2}e^{-\sqrt{\frac{10}{3} }z}[/tex]
After solving both equations, we have;[tex]C_{A}=\frac{e^{\sqrt{\frac{10}{3} } z}}{10}((1-\frac{e^{\sqrt{\frac{10}{3} }z}}{{-e^{\sqrt{\frac{10}{3} }z}}+{e^{-\sqrt{\frac{10}{3} }z}}} ) + (\frac{e^{\sqrt{\frac{10}{3} }z}}{{-e^{\sqrt{\frac{10}{3} }z}}+{e^{-\sqrt{\frac{10}{3} }z}}}))[/tex]
The we see that we can cancel terms, so we have the final solution that is:
[tex]C_{A}=\frac{e^{\sqrt{\frac{10}{3} }z}}{10}[/tex]
