Question

transform the given initial value problem into an equivalent problem with the initial point at the origin. $$ d y / d t=t^{2}+y^{2}, \quad y(1)=2 $$

Short answer

Answer: The equation of the equivalent problem with the initial point at the origin is $\frac{dy'}{dt'} = (t'+1)^2 + (y'-2)^2$ with the initial condition $y'(0) = 0$.

Step-by-step solution

Introduce a change of variables

Let \(t' = t-1\) and \(y' = y+c\). These new variables are obtained by shifting the time variable and the dependent variable.

Rewrite the given initial condition

Rewrite the given initial condition using the change of variables: $$(t',y')=(0, 2+c),$$ since when \(t = 1\), \(t'=0\), and when \(y=2\), \(y'= 2+c\).

Determine the transformed initial condition

Since the goal is to get the transformed initial condition to \(y(0)=0\), we need to solve \(2+c=0 \Rightarrow c=-2\). So, the transformed initial condition is: $$y'(-1) = 0.$$

Rewrite the given differential equation

Rewrite the given differential equation, \(\frac{dy}{dt} = t^2 + y^2\), using the new variables, \(t'\) and \(y'\). To do this, first we need to compute the derivative \(\frac{dy'}{dt'}\) by accounting for the relationship between the new variables \(t'\) and \(y'\), and the original \(t\) and \(y\). By the chain rule from calculus, we have: $$\frac{dy'}{dt'} = \frac{dy'}{dt} \times \frac{dt}{dt'}.$$ Since \(\frac{dy'}{dt} = \frac{dy}{dt}\) and \(\frac{dt}{dt'} = 1\), we get: $$\frac{dy'}{dt'} = \frac{dy}{dt} = t^2 + y^2 = (t'+1)^2 + (y'-2)^2,$$ where we have substituted \(t = t'+1\) and \(y = y'+2\).

Conclude the equivalent problem with the transformed variables

Finally, the equivalent problem with the initial point at the origin is the following initial value problem:$$\frac{dy'}{dt'} = (t'+1)^2 + (y'-2)^2,\quad y'(0) = 0.$$

Key concepts

Understanding Differential Equations

Differential equations are a type of equation that involves functions and their derivatives. They express a relationship between a function and its rate of change. In physics, engineering, economics, and various other fields, differential equations play a crucial role in modeling the behavior of dynamic systems.

In the given exercise, the differential equation \( dy / dt = t^2 + y^2 \) models how the function \( y(t) \) changes over time \( t \). Solving such an equation involves finding a function \( y(t) \) that satisfies this relationship for the given conditions. These conditions are provided by the initial value, which in this case is \( y(1) = 2 \).

The Role of Change of Variables

Applying a change of variables in differential equations can simplify the problem or provide a different perspective on the solution. The change of variables technique often involves introducing new variables to replace the old ones with the aim of making the equation more manageable or setting it in a preferred form.

In our exercise, the substitution \( t' = t-1 \) and \( y' = y+c \) shifts the graph of the solution so that the new initial conditions are at the origin. This makes it easier to understand and solve the problem. Steps need to be taken to ensure that this change of variables accurately reflects the original equation's behavior in the new variables. The transformation must be applied to both the differential equation and the initial conditions to maintain the equivalence of the problem.

Transforming Initial Conditions

Transforming initial conditions in differential equations involves modifying the starting values of the problem to suit the new variables. This transformation is a critical step that ensures the solution to the differential equation corresponds accurately under the change of variables.

The solution \( y'(0) = 0 \) is achieved by setting the constant \(c\) in our change of variables to be \( -2 \), based on the equation \( 2+c=0 \). The transformed differential equation is then expressed in terms of the new variables \( t' \) and \( y' \) with the initial condition \( y'(0) = 0 \) at the origin. This approach tailors the initial value problem to a more standard or simpler form, facilitating easier methods of finding the solution or applying further analytical techniques.