91Ó°ÊÓ

The given initial value problem is: $$ \frac{dy}{dt} = 1 - y^3, \quad y(-1) = 3 $$

Let's first find the general solution of the given differential equation, ignoring the initial condition for now. To do this, we need to solve: $$ \frac{dy}{dt} = 1 - y^3 $$ This is a first-order separable differential equation. To solve it, we will separate the variables and then integrate both sides. Separate the variables as follows: $$ \frac{dy}{1 - y^3} = dt $$ Now integrate both sides with respect to their respective variables: $$ \int \frac{dy}{1 - y^3} = \int dt $$ Integrating gives us: $$ \frac{1}{\sqrt[3]{2}}\operatorname{arctanh}\left(\frac{y}{\sqrt[3]{2}}\right) = t + C $$ where C is the constant of integration.

Now we will use the initial condition given in the problem \(y(-1) = 3\). Plugging this into our general solution, we get: $$ \frac{1}{\sqrt[3]{2}}\operatorname{arctanh}\left(\frac{3}{\sqrt[3]{2}}\right) = -1 + C $$ Solving for C, we get: $$ C = \frac{1}{\sqrt[3]{2}}\operatorname{arctanh}\left(\frac{3}{\sqrt[3]{2}}\right) + 1 $$

Substituting C into our solution from Step 2, the complete solution for our given initial value problem becomes: $$ \frac{1}{\sqrt[3]{2}}\operatorname{arctanh}\left(\frac{y}{\sqrt[3]{2}}\right) = t + \frac{1}{\sqrt[3]{2}}\operatorname{arctanh}\left(\frac{3}{\sqrt[3]{2}}\right) + 1 $$

Let's denote the transformed function as \(y_T(t)\) and the initial value of this function as \(y_T(0) = y_0\). We want to find a function \(y_T(t)\) that has the same shape as our original solution but is shifted so that its initial point is at the origin. To do this, we can simply replace \(t\) with \((t-T)\) in our solution and use the new initial condition \(y_T(0) = y_0\) to find \(T\): $$ \frac{1}{\sqrt[3]{2}}\operatorname{arctanh}\left(\frac{y_T}{\sqrt[3]{2}}\right) = (t - T) + \frac{1}{\sqrt[3]{2}}\operatorname{arctanh}\left(\frac{3}{\sqrt[3]{2}}\right) + 1 $$ Now, plug in the new initial condition \(y_T(0) = y_0\): $$ \frac{1}{\sqrt[3]{2}}\operatorname{arctanh}\left(\frac{y_0}{\sqrt[3]{2}}\right) = - T + \frac{1}{\sqrt[3]{2}}\operatorname{arctanh}\left(\frac{3}{\sqrt[3]{2}}\right) + 1 $$ Solve for \(T\): $$ T = \frac{1}{\sqrt[3]{2}}\operatorname{arctanh}\left(\frac{3}{\sqrt[3]{2}}\right) - \frac{1}{\sqrt[3]{2}}\operatorname{arctanh}\left(\frac{y_0}{\sqrt[3]{2}}\right) + 1 $$

The equivalent problem with the initial point at the origin is: $$ \frac{d y_T}{dt} = 1 - {y_T}^3, \quad y_T(0) = y_0 $$ and the transformed solution is: $$ \frac{1}{\sqrt[3]{2}}\operatorname{arctanh}\left(\frac{y_T}{\sqrt[3]{2}}\right) = (t - T) + \frac{1}{\sqrt[3]{2}}\operatorname{arctanh}\left(\frac{3}{\sqrt[3]{2}}\right) + 1 $$ with $$ T = \frac{1}{\sqrt[3]{2}}\operatorname{arctanh}\left(\frac{3}{\sqrt[3]{2}}\right) - \frac{1}{\sqrt[3]{2}}\operatorname{arctanh}\left(\frac{y_0}{\sqrt[3]{2}}\right) + 1 $$