Question

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

y'= 2y(1-y/4). y(0)=1


a. Find an Implicit expression of all solutions y of the differential equation above, in the form Ψ(t, y)= c, where c collects all constant terms. (So, do not include any c in your answer.)

Ψ(t, y) = ____________

Hint: Recall the partial fraction decomposition: k/y(k-y)= 1/y+ 1/(k-y), where k is any constant.

b. Find the explicit expression of the solution y of the initial value problem above.

y(t)=________

Answer 1

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-(y)/(4) \right )\\\frac{\mathrm{d} y}{\mathrm{d} t}=(y)/(2)(4-y)`

On integrating both sides, we get

`\int (dy)/(y(4-y))=\int (1)/(2)\,dt\\(1)/(4)\int (1)/(y)+(1)/(4-y)\,dy=(1)/(2)\,dt\\`

We know that
`\int (dy)/(y)=\ln y`.

Therefore,

`\int (dy)/(y(4-y))=\int (1)/(2)\,dt\\(1)/(4)\int (1)/(y)+(1)/(4-y)\,dy=\int (1)/(2)\,dt\\(1)/(4)\left [ \ln y-\ln (4-y) \right ] =(t)/(2)+C`

As
`y(0)=1`,

`C=-(\ln 3)/(4)`

So,
`(1)/(4)\left [ \ln y-\ln (4-y) \right ] =(t)/(2)-(\ln 3)/(4)`

b.

`(1)/(4)\left [ \ln y-\ln (4-y) \right ] =(t)/(2)-(\ln 3)/(4)\\\ln y-\ln (4-y)=2t-\ln 3\\\ln \left ( (y)/(4-y) \right )=2t-\ln 3\\(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=(4e^(2t-\ln 3))/(1+e^(2t-\ln 3))`