Question

(a) Solve the initial value problem by (i) transforming the given Bernoulli differential equation and initial condition into a first order linear differential equation with its corresponding initial condition, (ii) solving the new initial value problem, (iii) transforming back to the dependent variable of interest. (b) Determine the interval of existence. 13\. \(y^{\prime}=y(2-y), \quad y(0)=1\)

Short answer

Question: Transform the given Bernoulli differential equation into a linear differential equation and solve it. Determine the interval of existence for the solution. The differential equation is \(y'(t)=y(t)(2-y(t))\), with the initial condition \(y(0)=1\). Answer: The solution to the transformed linear differential equation is \(y(t)=\sqrt{2-2e^{-2t}}\), and the interval of existence is \([0, \infty)\).

Step-by-step solution

Transform the Bernoulli differential equation into a first-order linear differential equation

The given differential equation is of the form: \(y^{\prime}=y(2-y)\) This is a Bernoulli differential equation with \(n=-1\). To transform this into a linear differential equation, we need to find a substitution of the form \(v=y^{1-n}\). In our case, \(n=-1\), so our substitution will be: \(v=y^{1-(-1)}=y^2\) Now we find the derivative of v with respect to t: \(\frac{dv}{dt}=2yy'\) Substitute the given differential equation: \(\frac{dv}{dt}=2y(y(2-y))\) Now, our linear differential equation becomes: \(\frac{dv}{dt}=4y^2-2y^3\) And our initial condition is: \(v(0)=y^2(0)=1^2=1\)

Solve the new initial value problem

To solve \(v'(t)=4v-2v^2\) with the initial condition \(v(0)=1\), we first separate the variables: \(\frac{dv}{4v-2v^2} = dt\) Integrate both sides: \(-\frac{1}{2}\log|4-2v|=t+C_1\) Now we have: \(\log|4-2v|=-2t+C_2\) Where \(C_2=2C_1\). To get rid of the logarithm, we can take the exponential of both sides: \(4-2v=e^{-2t+C_2}\) At this point, we can satisfy the initial condition \(v(0)=1\): \(4-2(1)=e^{-2(0)+C_2}\) We find \(C_2 = \log(2)\), so, \(4-2v=2e^{-2t}\)

Transform back to the dependent variable of interest

To find y(t), we need to express y in terms of v, recall that \(v=y^2\): \(y(t)=\sqrt{v(t)}\) Now, solve for v: \(v(t)=2-2e^{-2t}\) And finally, find y(t): \(y(t)=\sqrt{2-2e^{-2t}}\)

Determine the interval of existence

Since the solution \(y(t)=\sqrt{2-2e^{-2t}}\) is defined for all \(t\geq 0\), the interval of existence is \([0, \infty)\).

Key concepts

Understanding Initial Value Problems

An initial value problem in mathematics involves finding the function that solves a differential equation and also satisfies a specific condition (the initial value) at a given point. For instance, with a first order differential equation such as our example problem, we're given an equation relating a function and its derivative, alongside an initial condition like \( y(0)=1 \). This condition acts as a starting point, anchoring our solution to one specific curve among the possible family of solutions.

For students, the challenge often lies in understanding the importance of this initial condition. It restricts the solution to one that not only solves the differential equation but also passes through a given point in the function's domain, providing a unique solution. Without it, we cannot determine a single, detailed trajectory for the function's behavior over time.

Tackling First Order Linear Differential Equations

In dealing with first order linear differential equations, remember that they have a standard form, which looks like \( y'+ p(x)y = q(x) \). Our Bernoulli equation, despite not immediately looking linear, can actually be transformed into this linear form through an ingenious substitution once we identify that it's a Bernoulli equation. This type of equation is indispensable in various scientific fields, as it can describe numerous natural phenomena, like population growth or decay processes.

To solve such equations, we employ methods like integrating factors or, as in our example, substitution which simplifies the equation into a linear form. It is crucial to be comfortable with these processes and recognize the standard forms of differential equations to apply the appropriate methods.

Exploring Interval of Existence

The interval of existence refers to the range of values, particularly in the independent variable (usually time), for which a solution to a differential equation is valid. In physical terms, it can be thought of as the period over which our mathematical model accurately represents the real system we are analyzing.

In the context of our exercise, after solving the differential equation, we ascertain this interval by looking at where the solution makes sense mathematically — for example, where it does not produce undefined or imaginary values. The interval \( [0, \(infty\)) \) for our solution implies the function remains valid for all positive times, a common scenario in processes that do not have an inherent expiration or singularities in the foreseeable future.

Separation of Variables Method

The separation of variables method is a staple in the toolbox for solving differential equations. It involves rearranging a differential equation so that each variable appears on a different side of the equation. This technique is applicable under certain conditions, especially when the equation is separable, meaning it can be manipulated into the form \( g(y)dy = f(x)dx \).

In our example problem, after transforming the initial Bernoulli equation into a linear form, we separated the \( v \) terms from the \( t \) terms before integrating. Mastery of this technique requires practice in algebraic manipulation and integral calculus. It is particularly useful because it transforms a potentially complex differential equation problem into a simpler integration problem.

Transforming Differential Equations

Sometimes, the direct approach to solving a differential equation can seem daunting. This is where transforming differential equations becomes a powerful strategy. In essence, we find a substitution that simplifies the original equation into a more manageable form. With the transformed equation solved, we then invert the substitution to return to the original variable.

Take our Bernoulli equation; the direct attack may not be straightforward. But by recognizing it as a Bernoulli and applying the suitable transformations, we arrived at a linear equation that could be dealt with comfortably. Both the initial transformation and the final reversion hinge on a strong grasp of algebra and the relationships between functions and their derivatives. This technique also elucidates powerful connections between different types of differential equations.