Chapter 9, Problem 1RE

Explain why or why not Determine whether the following statements are true and give an explanation or counterexample.

a.    The differential equation y′ + 2y = t is first-order, linear, and separable.

b.    The differential equation y′y = 2t² is first-order, linear, and separable.

c.    The function y = t + 1/t satisfies the initial value problem ty′ + y = 2t, y(1) = 2.

d.    The direction field for the differential equation y′(t) = t + y(t) is plotted in the ty-plane.

e.    Euler’s method gives the exact solution to the initial value problem y′ = ty², y(0) = 3 on the interval [0, a] provided a is not too large.

To determine

Whether the given statement is true or false.

Answer to Problem 1RE

The statement is False.

Explanation of Solution

The statement is the differential equation is first order, linear, and separable.

The given equation is $y^{\prime }\left(t\right)+2y=t$.

The order of this equation is one.

Thus, the equation is in first order.

The function y and its derivatives are in first order and not composed with other functions.

A linear equation cannot have products or quotients of y and its derivatives.

Thus, the equation is linear.

In the equation $g\left(y\right)y^{\prime }\left(t\right)=h\left(t\right)$, the factor $g\left(y\right)$ involves only y, and $h\left(t\right)$ involves only t, that is, the variables have been separated.

But here the equation is not separable.

Therefore, the equation is in first order, linear but not separable.

Thus, the statement is false.

To determine

Whether the given statement is true or false.

Answer to Problem 1RE

The statement is False.

Explanation of Solution

The given equation is $y^{\prime }y=2t^2$.

The order of this equation is one.

Thus, the equation is in first order.

Here, the function y and its derivatives are in first order and not composed with other functions.

A linear equation cannot have products or quotients of y and its derivatives.

Thus, the equation is not linear.

In the equation $g\left(y\right)y^{\prime }\left(t\right)=h\left(t\right)$, the factor $g\left(y\right)$ involves only y, and $h\left(t\right)$ involves only t, that is ,the variables have been separated.

But here the equation is separable.

Therefore, the equation is in first order, separable but not linear.

Thus, the statement is False.

To determine

Whether the given statement is true or false.

Answer to Problem 1RE

The statement is true.

Explanation of Solution

The given differential equation is $y=t+\frac{1}{t}$.

The initial value problem is $t{y}^{\prime }+y=2t,y\left(1\right)=2$.

Take derivative on both sides in $y=t+\frac{1}{t}$,

$\begin{array}{l}y=t+\frac{1}{t}\\ y^{\prime }=1-t^2\end{array}$

Substitute the value of $y^{\prime }$ in the initial value problem,

$\begin{array}{l}t{y}^{\prime }+y=2t\\ t\left(1-t^2\right)+y=2t\\ t-t^{-1}+\left(t+t^{-1}\right)=2t\\ 2t=2t\end{array}$

Therefore, the function $y=t+\frac{1}{t}$ satisfies the initial value problem $t{y}^{\prime }+y=2t$.

Thus, the statement is true.

To determine

Whether the direction field for the differential equation $y^{\prime }\left(t\right)=t+y\left(t\right)$ is plotted in the $ty$-plane.

Answer to Problem 1RE

True, the direction field for the differential equation $y^{\prime }\left(t\right)=t+y\left(t\right)$ is plotted in the $ty$-plane.

Explanation of Solution

The differential equation is $y^{\prime }\left(t\right)=t+y\left(t\right)$.

The equation is in first order

So, the notation $f\left(t,y\right)$ is an expression involving the independent variable t and the unknown solution y.

The differential equation at each point $\left(t,y\right)$ of the solution curve, the slope of the curve is $y^{\prime }\left(t\right)=t+y\left(t\right)$.

A direction field is a picture that shows the slope of the solution at $ty$-plane.

Therefore the direction field for the differential equation $y^{\prime }\left(t\right)=t+y\left(t\right)$ is plotted in the $ty$-plane.

To determine

Whether the given statement is true or false.

Answer to Problem 1RE

The statement is false.

Explanation of Solution

The carry capacity either greater than or less than the value predicted by the model.

The given initial value problem is $y^{\prime }=t{y}^2,y\left(0\right)=3$ on the interval $\left[0,a\right]$, where a is not too large.

Direction fields are the basis for many computer based methods for approximating solutions of a differential equation.

The exact solution of the initial value problem at grid points is $y\left(t_k\right)$, for $k=0,1,\mathrm{2....}N$.

Which is generally unknown unless solve the original differential equation.

The goal is to compute a set of approximations to the exact solution at the grid points.

Therefore, the given assumption is false Euler method gives approximate solution not exact solution.

Thus, the statement is false.