Chapter 1

Differential Equations and Linear Algebra ยท 324 exercises

Problem 50

Verify that the given function is a solution to the given differential equation. In these problems, \(c_{1}\) and \(c_{2}\) are arbitrary constants. $$\begin{aligned} &\diamond y(x)=\sum_{k=0}^{10} \frac{1}{k !} x^{k}, x y^{\prime \prime}-(x+10) y^{\prime}+10 y=0\\\ &x>0 \end{aligned}$$.

4 step solution

Problem 51

Solve the given initial-value problem. $$\frac{d y}{d x}+\frac{2 x}{1+x^{2}} y=x y^{2}, \quad y(0)=1$$

3 step solution

Problem 51

The velocity (meters/second) of an object at time \(t\) (seconds) is governed by the differential equation $$ \frac{d v}{d t}+k v=80 k e^{-k t} $$ with initial conditions $$ v(0)=20, \quad \frac{d v}{d t}(0)=2 $$ (a) How is the velocity of the object changing at \(t=0 ?\) (b) Determine the value of the constant \(k\) (c) Determine the velocity of the object at time \(t\) (d) Is there a finite time \(t>0\) at which the object is at rest? Explain. (e) What happens to the velocity of the object as \(t \rightarrow \infty ?\)

5 step solution

Problem 52

Solve the given initial-value problem. $$y^{\prime}+y \cot x=y^{3} \sin ^{3} x, \quad y(\pi / 2)=1$$

6 step solution

Problem 52

The temperature of an object at time \(t\) is governed by the linear differential equation $$ \frac{d T}{d t}=-k(T-5 \cos 2 t) $$At \(t=0,\) the temperature of the object is \(0^{\circ} \mathrm{F}\) and is, at that time, increasing at a rate of \(5^{\circ} \mathrm{F} / \mathrm{min.}\) (a) Determine the value of the constant \(k\) (b) Determine the temperature of the object at time \(t\) (c) Describe the behavior of the temperature of the object for large values of \(t\)

3 step solution

Problem 52

\diamondOne solution to the Bessel equation of (nonnegative) integer order \(N\) $$x^{2} y^{\prime \prime}+x y^{\prime}+\left(x^{2}-N^{2}\right) y=0$$ is $$y(x)=J_{N}(x)=\sum_{k=0}^{\infty} \frac{(-1)^{k}}{k !(N+k) !}\left(\frac{x}{2}\right)^{2 k+N}$$ (a) Write the first three terms of \(J_{0}(x)\) (b) Let \(J(0, x, m)\) denote the \(m\) th partial sum $$ J(0, x, m)=\sum_{k=0}^{m} \frac{(-1)^{k}}{(k !)^{2}}\left(\frac{x}{2}\right)^{2 k} $$ Plot \(J(0, x, 4)\) and use your plot to approximate the first positive zero of \(J_{0}(x) .\) Compare your value against a tabulated value or one generated by a computer algebra system. (c) Plot \(J_{0}(x)\) and \(J(0, x, 4)\) on the same axes over the interval \([0,2] .\) How well do they compare? (d) If your system has built-in Bessel functions, plot \(J_{0}(x)\) and \(J(0, x, m)\) on the same axes over the interval [0,10] for various values of \(m .\) What is the smallest value of \(m\) that gives an accurate approximation to the first three positive zeros of \(J_{0}(x) ?\)

4 step solution

Problem 53

Consider the differential equation $$y^{\prime}=F(a x+b y+c)$$ Where \(a, b \neq 0,\) and \(c\) are constants. Show that the change of variables from \(x, y\) to \(x, V,\) where $$V=a x+b y+c$$ reduces Equation (1.8.17) to the separable form $$ \frac{1}{b F(V)+a} d V=d x $$

4 step solution

Problem 53

Each spring, sandhill cranes migrate through the Platte River valley in central Nebraska. An estimated maximum of a half-million of these birds reach the region by April 1 each year. If there are only 100,000 sandhill cranes 15 days later and the sandhill cranes leave the Platte River valley at a rate proportional to the number of sandhill cranes still in the valley at the time, (a) How many sandhill cranes remain in the valley 30 days after April \(1 ?\) (b) How many sandhill cranes remain in the valley 35 days after April \(1 ?\) (c) How many days after April 1 will there be less than 1000 sandhill cranes in the valley?

4 step solution

Problem 54

Use the result from the previous problem to solve the given differential equation. For Problem 54 impose the given initial condition as well. $$y^{\prime}=(9 x-y)^{2}, \quad y(0)=0$$

6 step solution

Problem 55

Consider an RC circuit with \(R=4 \Omega, C=\frac{1}{5} \mathrm{F},\) and \(E(t)=6 \cos 2 t\) V. If \(q(0)=3 \mathrm{C},\) determine the current in the circuit for \(t \geq 0\)

4 step solution

Problem 56

Consider the RL circuit with \(R=3 \Omega, L=0.3 \mathrm{H},\) and \(E(t)=10 \mathrm{V} .\) If \(i(0)=3 \mathrm{A},\) determine the current in the circuit for \(t \geq 0\)

5 step solution

Problem 57

Show that the change of variables \(V=x y\) transforms the differential equation $$ \frac{d y}{d x}=\frac{y}{x} F(x y) $$ into the separable differential equation $$ \frac{1}{V[F(V)+1]} \frac{d V}{d x}=\frac{1}{x} $$

4 step solution

Problem 57

A solution containing 3 grams/L of a salt solution pours into a tank, initially half full of water, at a rate of 6 L/min. The well-stirred mixture flows out at a rate of \(4 \mathrm{L} / \mathrm{min}\). If the tank holds \(60 \mathrm{L}\), find the amount of salt (in grams) in the tank when the solution overflows.

3 step solution

Problem 58

Use Euler's method with the specified step size to determine the solution to the given initial-value problem at the specified point. $$y^{\prime}=x^{2}+2 y^{2}, \quad y(0)=-3, \quad h=0.1, \quad y(1)$$

4 step solution

Problem 59

Consider the differential equation $$ \frac{d y}{d x}=2 x(x+y)^{2}-1 $$ (a) Show that the change of variables $$ y(x)=w(x)-x $$ reduces \((1.8 .18)\) to the separable equation $$ \frac{d w}{d x}=2 x w^{2} $$ (b) Solve the differential equation ( 1.8 .19 ) and then determine the particular solution to the differential equation \((1.8 .18)\) that satisfies the initial condition \(y(0)=1\)

5 step solution

Problem 59

Use Euler's method with the specified step size to determine the solution to the given initial-value problem at the specified point. $$y^{\prime}=\frac{3 x}{y}+2, \quad y(1)=2, \quad h=0.05, \quad y(1.5)$$

4 step solution

Problem 60

Consider the differential equation $$ \frac{d y}{d x}=\frac{x+2 y-1}{2 x-y+3} $$ (a) Show that the change of variables defined by $$ x=u-1, \quad y=v+1 $$ transforms Equation (1.8.20) into the homogeneous equation $$ \frac{d v}{d u}=\frac{u+2 v}{2 u-v} $$ (b) Find the general solution to Equation and hence, solve Equation ( 1.8 .20 ).

5 step solution

Problem 61

A differential equation of the form $$ y^{\prime}+p(x) y+q(x) y^{2}=r(x) $$ is called a Riccati equation. (a) If \(y=Y(x)\) is a known solution to Equation \((1.8 .22),\) show that the substitution $$ y=Y(x)+v^{-1}(x) $$ reduces it to the linear equation $$ v^{\prime}-[p(x)+2 Y(x) q(x)] v=q(x) $$ (b) Find the general solution to the Riccati equation $$ x^{2} y^{\prime}-x y-x^{2} y^{2}=1, \quad x>0 $$ given that \(y=-x^{-1}\) is a solution.

3 step solution

Problem 62

Consider the Riccati equation $$ y^{\prime}+2 x^{-1} y-y^{2}=-2 x^{-2}, \quad x>0 $$ (a) Determine the values of the constants \(a\) and \(r\) such that \(y(x)=a x^{r}\) is a solution to Equation \((1.8 .23)\) (b) Use the result from part (a) of the previous problem to determine the general solution to Equation \((1.8 .23)\)

3 step solution

Problem 63

(a) Show that the change of variables \(y=x^{-1}+w\) transforms the Riccati differential equation $$ y^{\prime}+7 x^{-1} y-3 y^{2}=3 x^{-2} $$ into the Bernoulli equation $$ w^{\prime}+x^{-1} w=3 w^{2} $$ (b) Solve Equation \((1.8 .25),\) and hence determine the general solution to ( 1.8 .24 ).

6 step solution

Problem 64

Consider the differential equation $$ y^{-1} y^{\prime}+p(x) \ln y=q(x) $$ where \(p(x)\) and \(q(x)\) are continuous functions on some interval \((a, b) .\) Show that the change of variables \(u=\ln y\) reduces Equation \((1.8 .26)\) to the linear differential equation $$ u^{\prime}+p(x) u=q(x) $$ and hence show that the general solution to Equation \((1.8 .26)\) is $$ y(x)=\exp \left\\{I^{-1}\left[\int I(x) q(x) d x+c\right]\right\\} $$ where $$ I=e^{\int p(x) d x} $$ and \(c\) is an arbitrary constant.

4 step solution

Problem 65

Use the technique derived in the previous problem to solve the initial-value problem $$ \begin{aligned} y^{-1} y^{\prime}-2 x^{-1} \ln y &=x^{-1}(1-2 \ln x) \\ y(1) &=e \end{aligned} $$

4 step solution

Problem 66

Consider the differential equation $$ f^{\prime}(y) \frac{d y}{d x}+p(x) f(y)=q(x) $$ where \(p\) and \(q\) are continuous functions on some interval \((a, b),\) and \(f\) is an invertible function. Show that Equation \((1.8 .28)\) can be written as $$ \frac{d u}{d x}+p(x) u=q(x) $$ where \(u=f(y)\) and hence show that the general solution to Equation \((1.8 .28)\) is $$ y(x)=f^{-1}\left\\{I^{-1}\left[\int I(x) q(x) d x+c\right]\right\\} $$ where \(I\) is given in \((1.8 .27), f^{-1}\) is the inverse of \(f\) and \(c\) is an arbitrary constant.

2 step solution

Problem 67

Solve $$ \sec ^{2} y \frac{d y}{d x}+\frac{1}{2 \sqrt{1+x}} \tan y=\frac{1}{2 \sqrt{1+x}} $$

4 step solution

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