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Chapter 2: Problem 2

In Problems \(1-6,\) determine whether the given differential equation is separable. $$ \frac{d y}{d x}=4 y^{2}-3 y+1 $$

Short Answer

Expert verified
The given differential equation is not separable.

Step by step solution

01

Check if the equation is separable

The differential equation given is \(\frac{d y}{d x}=4 y^{2}-3 y+1\). A first glance at the problem shows that the right side of the given equation can't be factored into a product of functions of y and x separately, thus it seems that the equation is not separable. However, we need to explore a little more to confirm this.
02

Try expressing the equation in the form \( f(y)dy = g(x)dx \)

An equation can't be called non-separable just by looking, there might be some way to rewrite it such that it becomes separable. However, in this case, there isn't any x term in the given equation, so we cannot express it as a function of x times dx equals a function of y times dy. It confirms our initial judgment that the equation is indeed non-separable.

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Key Concepts

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

Ordinary Differential Equations (ODEs)
Ordinary differential equations (ODEs) are equations that involve functions and their derivatives.
These equations express the relationship between a function and its rate of change. ODEs are crucial in modeling various phenomena in science and engineering, such as motion, heat, and fluid flow.
  • **Order of an ODE:** The order of an ordinary differential equation is determined by the highest derivative present. If the highest derivative is the first derivative, it's a first-order ODE; if the highest is the second derivative, it's a second-order ODE, and so on.
  • **Linear vs. Nonlinear:** ODEs can be linear or nonlinear. Linear ODEs have solutions that can be expressed as linear combinations of the unknown functions and their derivatives, whereas nonlinear ODEs involve products or powers of the unknown functions.
  • **Initial Value Problems (IVP):** An IVP for an ODE includes conditions specified at a single point, which helps determine a unique solution among many potential ones.
Understanding ODEs is essential for interpreting the behavior of dynamic systems and for effectively applying mathematical models to real-world problems.
Factoring Expressions
Factoring is a mathematical process used to break down expressions into simpler components, or 'factors,' that, when multiplied together, give the original expression.
This technique is often used to simplify problems and make them easier to solve. It plays a crucial role in algebra and calculus related tasks.
  • **Common Factoring Techniques:** Factoring can involve methods like pulling out a greatest common factor (GCF), applying the difference of squares, or factoring trinomials into binomial products.
  • **Importance in Differential Equations:** In the context of differential equations, factoring can help to transform equations into a solvable format, particularly when looking for solutions to linear equations or seeking to determine if an equation is separable.
While the process of factoring is straightforward in many cases, some expressions may be complex enough to appear non-factorable at first glance, necessitating deeper analysis or advanced techniques.
Non-separable Equations
Non-separable differential equations are those that cannot be written in the form where each variable can be placed on separate sides of the equation, like \(f(y)dy = g(x)dx\).
In other words, they cannot be expressed as the product of two functions, one depending solely on \(x\) and the other solely on \(y\).
  • **Characteristics:** These equations contain terms that mix derivatives and the function itself in a way that precludes simple separation. Non-separable equations challenge simple integration techniques, often requiring more complex methods for solving, such as numerical approaches or different representation strategies.
  • **Why Non-separable?:** In our example, \(\frac{dy}{dx}=4y^2-3y+1\), there's no straightforward way to split the terms regarding \(x\) and \(y\). It's not possible to arrange the expression into separate multiplicative parts involving \(dx\) and \(dy\).
These equations highlight situations where interrelations between variables are too intertwined for straightforward separation, needing more intricate problem-solving strategies.

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Most popular questions from this chapter

Mixing. Suppose a brine containing 0.3 kilogram (kg) of salt per liter (L) runs into a tank initially filled with 400 L of water containing 2 kg of salt. If the brine enters at 10 L/min, the mixture is kept uniform by stirring, and the mixture flows out at the same rate. Find the mass of salt in the tank after 10 min (see Figure 2.4). [Hint: Let A denote the number of kilograms of salt in the tank at t min after the process begins and use the fact that rate of increase in $$A =$$ rate of input - rate of exit. A further discussion of mixing problems is given in Section 3.2.]

As stated in this section, the separation of equation (2) on page 42 requires division by \( p(y) \), and this may disguise the fact that the roots of the equation \( p(y)=0 \) are actually constant solutions to the differential equation. (a) To explore this further, separate the equation $$ \frac{d y}{d x}=(x-3)(y+1)^{2 / 3} $$ to derive the solution, $$ y=-1+\left(x^{2} / 6-x+C\right)^{3} $$ (b) Show that \( y \equiv-1 \) satisfies the original equation \( d y / d x=(x-3)(y+1)^{2 / 3} \) (c) Show that there is no choice of the constant \( C \) that will make the solution in part (a) yield the solution \( y \equiv-1 \) Thus, we lost the solution \( y=-1 \) when we divided by \((y+1)^{2 / 3}\).

Consider the equation $$\left(y^{2}+2 x y\right) d x-x^{2} d y=0$$ (a) Show that this equation is not exact. (b) Show that multiplying both sides of the equation by $$y^{-2}$$ yields a new equation that is exact. (c) Use the solution of the resulting exact equation to solve the original equation. (d) Were any solutions lost in the process?

Riccati Equation. An equation of the form $$\frac{d y}{d x}=P(x) y^{2}+Q(x) y+R(x)$$ is called a generalized Riccati equation. (a) If one solution_say,$$u(x)-\text { of }(18)$$ is known, show that the substitution $$y=u+1 / v \text { reduces }(18)$$ to a linear equation in v. (b) Given that $$u(x)=x$$ is a solution to $$\frac{d y}{d x}=x^{3}(y-x)^{2}+\frac{y}{x}$$ use the result of part (a) to find all the other solutions to this equation. (The particular solution $$u(x)=x$$ can be found by inspection or by using a Taylor series method; see Section 8.1.)

As discussed in calculus, certain indefinite integrals (antiderivatives) such as \( \int e^{x^{x}} d x \) cannot be expressed in finite terms using elementary functions. When such an integral is encountered while solving a differential equation, it is often helpful to use definite integration (integrals with variable upper limit). For example, consider the initial value problem $$ \frac{d y}{d x}=e^{x^{2}} y^{2}, \quad y(2)=1 $$ The differential equation separates if we divide by \( y^{2} \) and multiply by \( d x \). We integrate the separated equation from \( x=2 \) to \( x=x_{1} \) and find $$ \int_{x=2}^{x=x_{1}} e^{x^{2}} d x=\int_{x=2}^{x=x_{1}} \frac{d y}{y^{2}} $$ $$ =-\frac{1}{y} \left| \begin{array}{l}{x=x_{1}} \\ {x=2}\end{array}\right. $$ $$ =-\frac{1}{y\left(x_{1}\right)}+\frac{1}{y(2)} $$ If we let \( t \) be the variable of integration and replace \( x_{1} \) by \( x \) and \( y(2) \) by 1, then we can express the solution to the initial value problem by $$ y(x)=\left(1-\int_{2}^{x} e^{t^{2}} d t\right)^{-1} $$ Use definite integration to find an explicit solution to the initial value problems in parts (a)-(c). (a) \( d y / d x=e^{x^{2}}, \quad y(0)=0 \) (b) \( d y / d x=e^{x^{2}} y^{-2}, \quad y(0)=1 \) (c) \( d y / d x=\sqrt{1+\sin x}\left(1+y^{2}\right), \quad y(0)=1 \) (d) Use a numerical integration algorithm (such as Simpson's rule, described in Appendix C) to approximate the solution to part (b) at \( x=0.5 \) to three decimal places.
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