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Determine whether the differential equation is linear. $$ \frac{d R}{d t}+t \cos R=e^{-t} $$

Short Answer

Expert verified
The differential equation is non-linear due to the \( \cos R \) term.

Step by step solution

01

Identify the Standard Form

A differential equation is linear if it can be written in the standard form \( a(t) \frac{d R}{d t} + b(t) R = c(t) \). In this case, we are given \( \frac{d R}{d t} + t \cos R = e^{-t} \).
02

Examine the Terms Involving R

The equation involves the term \( t \cos R \), which is a non-linear function of \( R \). For a differential equation to be linear, \( R \) and its derivatives must only appear to the first power and must not be multiplied or composed with other functions.
03

Determine Linearity

Since \( \cos R \) is a non-linear function of \( R \), the given equation \( \frac{d R}{d t} + t \cos R = e^{-t} \) cannot be transformed into a form where \( R \) and its derivatives appear only to the first power. Therefore, it is a non-linear differential equation.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Linearity of Differential Equations
A differential equation is deemed linear if it satisfies certain characteristics, specifically in how the dependent variable and its derivatives are presented. In a linear differential equation:
  • The dependent variable and its derivatives appear only in the first power.
  • They do not multiply each other or interact with one another in functions like sine, cosine, or any other trigonometrical, exponential, or log functions.
When the equation is represented in a specific canonical form like: \[ a(t) \frac{dR}{dt} + b(t)R = c(t) \] the equation remains linear only if the coefficient terms \( a(t) \), \( b(t) \), and \( c(t) \) do not involve the dependent variable or its derivatives in complex forms. Thus, it's critical to scrutinize each term surrounding the dependent variable to ensure they meet the criteria for linearity.
Standard Form of Differential Equations
The standard form of a linear differential equation is an essential representation that allows us to observe and utilize the properties of linearity effectively. For first-order linear differential equations, the standard form is:
  • \( a(t) \frac{dR}{dt} + b(t)R = c(t) \)
Here, \( a(t) \), \( b(t) \), and \( c(t) \) are functions of the independent variable, usually time \( t \), and they act as coefficients.
Understanding this allows for simplifications and techniques to isolate \( R \), such as using the integrating factor or other methods.
Notably, checking an equation against the standard form helps establish whether it's linear. If an equation involves terms like \( t\cos R \) as seen in the example given, it doesn't fit the standard form, indicating non-linearity.
Non-linear Differential Equations
Non-linear differential equations are those where the dependent variable or its derivatives exist in a form that extends beyond first powers. Here are the key aspects:
  • They can involve powers greater than one, products, or other functions (like sine, cosine, or exponential) interacting with the dependent variable.
  • Non-linearity usually makes such equations more complex to solve, often requiring numerical methods or approximations rather than analytical solutions.
Unlike linear differential equations, there isn't a robust, universal methodology for solving all non-linear equations. Instead, solutions often depend on the specific equation itself, and bespoke strategies or numerical techniques might be necessary.
Recognizing non-linearity, by identifying terms that do not comply with the linear form, is key in deciding the approach to solving and understanding the equation at hand.

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Most popular questions from this chapter

A sphere with radius \(1 \mathrm{m}\) has temperature \(15^{\circ} \mathrm{C}\). It lies inside a concentric sphere with radius \(2 \mathrm{m}\) and temperature \(25^{\circ} \mathrm{C}\). The temperature \(T(r)\) at a distance \(r\) from the common center of the spheres satisfies the differential equation \(\frac{d^{2} T}{d r^{2}}+\frac{2}{r} \frac{d T}{d r}=0\) If we let \(S=d T / d r,\) then \(S\) satisfies a first-order differential equation. Solve it to find an expression for the temperature \(T(r)\) between the spheres.

Solve the differential equation. $$ \frac{d z}{d t}+e^{t+z}=0 $$

Find the solution of the differential equation that satisfies the given initial condition. $$ x+3 y^{2} \sqrt{x^{2}+1} \frac{d y}{d x}=0, \quad y(0)=1 $$

To account for seasonal variation in the logistic differential equation, we could allow \(k\) and \(M\) to be functions of \(t:\) $$ \frac{d P}{d t}=k(t) P\left(1-\frac{P}{M(t)}\right) $$ (a) Verify that the substitution \(z=1 / P\) transforms this equation into the linear equation $$ \frac{d z}{d t}+k(t) z=\frac{k(t)}{M(t)} $$ (b) Write an expression for the solution of the linear equation in part (a) and use it to show that if the carrying capacity \(M\) is constant, then $$ P(t)=\frac{M}{1+C M e^{-j k(t) d t}} $$ Deduce that if \(\int_{0}^{\infty} k(t) d t=\infty,\) then \(\lim _{t \rightarrow \infty} P(t)=M .\) IThis will be true if \(k(t)=k_{0}+a \cos b t\) with \(k_{0}>0,\) which describes a positive intrinsic growth rate with a periodic seasonal variation. (c) If \(k\) is constant but \(M\) varies, show that $$ z(t)=e^{-k t} \int_{0}^{t} \frac{k e^{k s}}{M(s)} d s+C e^{-k t} $$ and use I' Hospital's Rule to deduce that if \(M(t)\) has a limit as \(t \rightarrow \infty\), then \(P(t)\) has the same limit.

Sketch the direction field of the differential equation. Then use it to sketch a solution curve that passes through the given point. $$ y^{\prime}=y+x y, \quad(0,1) $$

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