Question

Consider the initial value problem for the function y given by:

Answer 1

Answer:

Step-by-step explanation:

a.

An implicit expression is a relation of the form y = f(x) where f is a a function with x as a variable.

$\frac{\mathrm{d} y}{\mathrm{d} t}=2y\left ( 1-\frac{y}{4} \right )\\\frac{\mathrm{d} y}{\mathrm{d} t}=\frac{y}{2}(4-y)$

On integrating both sides, we get

$\int \frac{dy}{y(4-y)}=\int \frac{1}{2}\,dt\\\frac{1}{4}\int \frac{1}{y}+\frac{1}{4-y}\,dy=\frac{1}{2}\,dt$

We know that $\int \frac{dy}{y}=\ln y$.

Therefore,

$\int \frac{dy}{y(4-y)}=\int \frac{1}{2}\,dt\\\frac{1}{4}\int \frac{1}{y}+\frac{1}{4-y}\,dy=\int \frac{1}{2}\,dt\\\frac{1}{4}\left [ \ln y-\ln (4-y) \right ] =\frac{t}{2}+C$

As $y(0)=1$,

$C=-\frac{\ln 3}{4}$

So, $\frac{1}{4}\left [ \ln y-\ln (4-y) \right ] =\frac{t}{2}-\frac{\ln 3}{4}$

b.

$\frac{1}{4}\left [ \ln y-\ln (4-y) \right ] =\frac{t}{2}-\frac{\ln 3}{4}\\\ln y-\ln (4-y)=2t-\ln 3\\\ln \left ( \frac{y}{4-y} \right )=2t-\ln 3\\\frac{y}{4-y}=e^{2t-\ln 3}\\y=(4-y)e^{2t-\ln 3}\\y\left ( 1+e^{2t-\ln 3}\\ \right )=4e^{2t-\ln 3}\\y=\frac{4e^{2t-\ln 3}}{1+e^{2t-\ln 3}}$