Question

(a) Show that \(\phi(t)=e^{2 t}\) is a solution of \(y^{\prime}-2 y^{\prime}-2 y=0\) and that \(y=c \phi(t)\) is also a solution of this equatio for any value of the conntanct \(y^{2}=0\) for \(t>0\), but that \(y=c \phi(t)\) is (b) Show that \(\phi(t)=1 / t\) is a solution of \(y^{\prime}+y^{2}=0\) for \(t>0\), but that \(y=c \phi(t)\) is not solution of this equation anless \(c=0\) or \(c=1 .\) Note that the equation of part (b) is nonlinear, while that of part (a) is linear.

Short answer

Question: Show that for the following differential equations, the given solutions are valid and find the general form. (a) Verify that \(y'=2y'-2y=0\) has solutions of the form \(y=c e^{2t}\), where c is any constant. (b) Verify that \(y'+y^2=0\) has a solution of the form \(y=\frac{1}{t}\) for \(t>0\), and find under which conditions there can be a solution of the form \(y=\frac{c}{t}\). Answer: (a) We have shown that \(y'=2y'-2y=0\) has a solution \(\phi(t)=e^{2t}\), and the general form of the solution is \(y=c e^{2t}\) for any constant c. (b) We have shown that \(y'+y^2=0\) has a solution \(\phi(t)=\frac{1}{t}\) for \(t>0\), and the general form of the solution is \(y=\frac{c}{t}\) with c only being 0 or 1.

Step-by-step solution

Part (a): Showing that \(\phi(t)=e^{2t}\) is a solution of the equation \(y'-2y'-2y=0\)

To show that \(\phi(t)=e^{2t}\) is a solution, we need to differentiate it with respect to \(t\) and plug it back into the equation. We have: \(\phi(t) = e^{2t}, \phi'(t) = 2e^{2t}\) Now, replace the \(y\) and its derivatives in the given differential equation by \(\phi(t)\) and \(\phi'(t)\), respectively: \(\phi'(t)-2\phi'(t)-2\phi(t) = 2e^{2t}-4e^{2t}-2e^{2t} = 0\) Since the equation holds, \(\phi(t)=e^{2t}\) is indeed a solution.

Part (a): Showing that \(y=c\phi(t)\) is a solution of the equation for any value of the constant

Now let's see if \(y=c\phi(t)\) is also a solution, where \(c\) is a constant. \(y(t) = c\phi(t) = ce^{2t}\) Now we will find the derivative: \(y'(t) = 2ce^{2t}\) Now, substitute this back into the equation: \(y'(t)-2y'(t)-2y(t) = 2ce^{2t}-4ce^{2t}-2ce^{2t} = 0\) We can observe that the equation holds for any value of the constant \(c\). Thus, \(y=c\phi(t)\) is also a solution of the given differential equation.

Part (b): Showing that \(\phi(t)=1/t\) is a solution of \(y'+y^2=0\) for \(t>0\)

To show that \(\phi(t)=1/t\) is a solution of the given equation, we need to differentiate it with respect to \(t\) and plug it back into the equation. We have: \(\phi(t) = 1/t, \phi'(t) = -1/t^2\) Now, let's plug these values into the equation: \(\phi'(t)+\phi^2(t) = -\frac{1}{t^2}+\frac{1}{t^2} = 0\) Since the equation holds for \(t>0\), \(\phi(t)=1/t\) is a solution of the given differential equation.

Part (b): Showing that \(y=c\phi(t)\) is not a solution for this equation unless \(c=0\) or \(c=1\)

Let's check if the scaling of this solution \(y(t)=c\phi(t)=c/t\) will also be a solution of this differential equation. Differentiating this expression with respect to \(t\) gives us: \(y'(t) = -c/t^2\) Now, let's plug this into the equation: \(y'(t) + y^2(t) = -\frac{c}{t^2} + \frac{c^2}{t^2} = \frac{c^2-c}{t^2}\) For this equation to hold, we only have two cases: 1. \(c^2-c=0 \Longrightarrow c(c-1)=0 \Longrightarrow c=0 \text{ or } c=1\) Thus, \(y=c\phi(t)\) is not a solution of this equation (given \(t>0\)) unless \(c=0\) or \(c=1\).

Key concepts

Linear Differential Equations

Linear differential equations have a defining characteristic: all terms involving the unknown function and its derivatives are linear. A linear differential equation takes the form \( a_n(t)\frac{d^n y}{dt^n} + a_{n-1}(t)\frac{d^{n-1} y}{dt^{n-1}} + \ldots + a_1(t)\frac{dy}{dt} + a_0(t)y = g(t) \) where each term is linear with respect to \( y \). In simpler terms, if we have unknowns like \( y \) or its derivatives, they always appear to the first power and are not multiplied by each other.
This makes the equations easier to solve, as solutions can often be found using integration or substitution. These types of equations often represent systems with a constant or time-varying proportionality and superposition. For the linear equation presented in the exercise, \( y' - 2y' - 2y = 0 \), substituting solutions like \( \phi(t) = ce^{2t} \) keeps the equation balanced for any constant \( c \).
Linear differential equations are common in physics and engineering, describing systems like circuits or oscillations.

Nonlinear Differential Equations

Nonlinear differential equations, in contrast to their linear counterparts, include terms that are nonlinear functions of the unknown function or its derivatives. This could be in the form of powers (other than first power), products of the function and its derivatives, or any other non-linear operation. The essence of a nonlinear differential equation can make them more complex and challenging to solve.
Consider the equation given in the exercise: \( y' + y^2 = 0 \). Unlike linear equations, the term \( y^2 \) introduces nonlinearity. Solving nonlinear problems often requires specific methods or numerical techniques and may not guarantee a closed-form solution.
These equations more accurately reflect real-world processes with complexities such as turbulence, chaotic systems, or biological phenomena where small changes can lead to large-scale effects. The specific nature of the solutions can depend heavily on initial conditions, leading to potential sensitivity or multiple solution branches.

Solution to Differential Equations

Finding a solution to a differential equation involves identifying a function (or a set of functions) that satisfies the equation across a specific domain. There are generally two types of solutions:

• Particular Solution: This satisfies the differential equation with specific initial and/or boundary conditions.
• General Solution: Represents all possible solutions to the differential equation and often involves arbitrary constants.

For example, for the linear differential equation in the exercise, \( y = c\phi(t) \) is the general solution, with \( \phi(t) = e^{2t} \). This solution works for any \( c \) value given the equation's form, showing its validity as a general solution for linear cases.
In contrast, for nonlinear equations like \( y' + y^2 = 0 \), the solution \( \phi(t) = \frac{1}{t} \) only holds unless modified by constants, leading to \( c \) being restricted to specific values. Thus, understanding how constants impact solutions is key to fully comprehending differential equations.

Differential Calculus

Differential calculus is a fundamental branch of mathematics that focuses on the concept of the derivative, which represents change. It's a critical tool for solving differential equations as it allows for understanding rates of change within a system.
When working with differential equations, calculus aids in:

• Finding derivatives: By determining the gradient of functions, which is integral in forming an equation.
• Initial and Boundary Conditions: Using derivatives to meet specified conditions at the start or endpoints shaping particular solutions.
• Integration: Provides the means to revert derivatives back to functions, revealing potential solutions.

In the exercise, differentiation allowed for forming the equation's expressions, checking potential solutions like \( \phi'(t) \), and integrating when necessary to find general solutions or specific conditions.
Overall, differential calculus is a foundational tool that offers the methodology needed to unravel differential equations, understanding how quantities dynamically evolve.