Linear Second-Order Equations

Find the solution to the initial value problem.$\mathbf{y}\mathbf{\text{'}}\mathbf{-}\mathbf{y}\mathbf{=}\mathbf{1}\mathbf{,}\mathbf{y}\left(\mathbf{0}\right)\mathbf{=}\mathbf{0}$

Find the solution to the initial value problem.

$\mathbf{y}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{=}\mathbf{6}\mathbf{t}\mathbf{;}\mathbf{y}\left(\mathbf{0}\right)\mathbf{=}\mathbf{3}\mathbf{,}\mathbf{y}\mathbf{\text{'}}\left(\mathbf{0}\right)\mathbf{=}\mathbf{-}\mathbf{1}$

Find the solution to the initial value problem.

$\mathbf{z}\mathbf{\text{'}}\mathbf{\text{'}}\left(\mathbf{x}\right)\mathbf{+}\mathbf{z}\left(\mathbf{x}\right)\mathbf{=}\mathbf{2}{\mathbf{e}}^{-x}\mathbf{,}\mathbf{z}\left(\mathbf{0}\right)\mathbf{=}\mathbf{0}\mathbf{,}\mathbf{z}\mathbf{\text{'}}\left(\mathbf{0}\right)\mathbf{=}\mathbf{0}$

Find the solution to the initial value problem.

$\mathbf{y}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{+}\mathbf{9}\mathbf{y}\mathbf{=}\mathbf{27}\mathbf{;}\mathbf{y}\left(\mathbf{0}\right)\mathbf{=}\mathbf{4}\mathbf{,}\mathbf{y}\mathbf{\text{'}}\left(\mathbf{0}\right)\mathbf{=}\mathbf{6}$

Find the solution to the initial value problem.

$\mathbf{y}\mathbf{\text{'}}\mathbf{\text{'}}\left(\mathbf{x}\right)\mathbf{-}\mathbf{y}\mathbf{\text{'}}\left(\mathbf{x}\right)\mathbf{-}\mathbf{2}\mathbf{y}\left(\mathbf{x}\right)\mathbf{=}\mathbf{cosx}\mathbf{-}\mathbf{sin}\mathbf{2}\mathbf{x}\mathbf{;}\mathbf{y}\left(\mathbf{0}\right)\mathbf{=}\frac{-7}{20}\mathbf{,}\mathbf{y}\mathbf{\text{'}}\left(\mathbf{0}\right)\mathbf{=}\frac{1}{5}$

Find the solution to the initial value problem.

$\mathbf{y}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{+}\mathbf{y}\mathbf{\text{'}}\mathbf{-}\mathbf{12}\mathbf{y}\mathbf{=}{\mathbf{e}}^t\mathbf{+}{\mathbf{e}}^{2t}\mathbf{-}\mathbf{1}\mathbf{;}\mathbf{y}\left(\mathbf{0}\right)\mathbf{=}\mathbf{1}\mathbf{,}\mathbf{y}\mathbf{\text{'}}\left(\mathbf{0}\right)\mathbf{=}\mathbf{3}$

Find the solution to the initial value problem.$\mathbf{y}\mathbf{\text{'}}\mathbf{\text{'}}\left(\mathbf{\theta }\right)\mathbf{-}\mathbf{y}\left(\mathbf{\theta }\right)\mathbf{=}\mathbf{sin\theta }\mathbf{-}{\mathbf{e}}^{2\theta }\mathbf{;}\mathbf{y}\left(\mathbf{0}\right)\mathbf{=}\mathbf{1}\mathbf{,}\mathbf{y}\mathbf{\text{'}}\left(\mathbf{0}\right)\mathbf{=}\mathbf{-}\mathbf{1}$

Find the solution to the initial value problem. $\mathbf{y}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{+}\mathbf{2}\mathbf{y}\mathbf{\text{'}}\mathbf{+}\mathbf{y}\mathbf{=}{\mathbf{t}}^2\mathbf{+}\mathbf{1}\mathbf{-}{\mathbf{e}}^t\mathbf{y}\left(\mathbf{0}\right)\mathbf{=}\mathbf{0}\mathbf{,}\mathbf{y}\mathbf{\text{'}}\left(\mathbf{0}\right)\mathbf{=}\mathbf{2}$

Determine the form of a particular solution for the differential equation. Do not solve.

$\mathit{x}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{-}\mathit{x}\mathbf{\text{'}}\mathbf{-}\mathbf{2}\mathit{x}\mathbf{=}{\mathit{e}}^{\mathbf{t}}\mathit{c}\mathit{o}\mathit{s}\mathit{t}\mathbf{-}{\mathit{t}}^{\mathbf{2}}\mathbf{+}\mathit{c}\mathit{o}{\mathit{s}}^{\mathbf{3}}\mathit{t}$

Determine the form of a particular solution for the differential equation. Do not solve

$\mathit{y}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{+}\mathbf{5}\mathit{y}\mathbf{\text{'}}\mathbf{+}\mathbf{6}\mathit{y}\mathbf{=}\mathit{s}\mathit{i}\mathit{n}\mathit{t}\mathbf{-}\mathit{c}\mathit{o}\mathit{s}\mathbf{2}\mathit{t}$

Determine the form of a particular solution for the differential equation. Do not solve

$\mathit{y}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{-}\mathbf{4}\mathit{y}\mathbf{\text{'}}\mathbf{+}\mathbf{5}\mathit{y}\mathbf{=}{\mathit{e}}^{\mathbf{5}\mathbf{t}}\mathbf{+}\mathit{t}\mathit{s}\mathit{i}\mathit{n}\mathbf{3}\mathit{t}\mathbf{-}\mathit{c}\mathit{o}\mathit{s}\mathbf{3}\mathit{t}$

Determine the form of a particular solution for the differential equation. Do not solve

$\mathit{y}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{-}\mathbf{4}\mathit{y}\mathbf{\text{'}}\mathbf{+}\mathbf{4}\mathit{y}\mathbf{=}{\mathit{t}}^{\mathbf{2}}{\mathit{e}}^{\mathbf{2}\mathit{t}}\mathbf{-}{\mathit{e}}^{\mathbf{2}\mathit{t}}$.

Find a particular solution to the given higher-order equation.

$\mathbf{y}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{-}\mathbf{2}\mathbf{y}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{-}\mathbf{y}\mathbf{\text{'}}\mathbf{+}\mathbf{2}\mathbf{y}\mathbf{=}\mathbf{2}{\mathbf{t}}^2\mathbf{+}\mathbf{4}\mathbf{t}\mathbf{-}\mathbf{9}$

Find a particular solution to the given higher-order equation.

${\mathbf{y}}^4\mathbf{-}\mathbf{5}\mathbf{y}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{+}\mathbf{4}\mathbf{y}\mathbf{=}\mathbf{10}\mathbf{cost}\mathbf{-}\mathbf{20}\mathbf{sint}$

Find a particular solution to the given higher-order equation. $\mathbf{y}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{+}\mathbf{y}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{-}\mathbf{2}\mathbf{y}\mathbf{=}{\mathbf{te}}^t\mathbf{+}\mathbf{1}$

Find a particular solution to the given higher-order equation.${\mathbf{y}}^4\mathbf{-}\mathbf{3}\mathbf{y}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{+}\mathbf{3}\mathbf{y}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{-}\mathbf{y}\mathbf{\text{'}}\mathbf{=}\mathbf{6}\mathbf{t}\mathbf{-}\mathbf{20}$

Discontinuous Forcing Term. In certain physical models, the nonhomogeneous term, or forcing term, g(t) in the equation

 $\mathbf{ay}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{+}\mathbf{by}\mathbf{\text{'}}\mathbf{+}\mathbf{cy}\mathbf{=}\mathbf{g}\left(\mathbf{t}\right)$

may not be continuous but may have a jump discontinuity. If this occurs, we can still obtain a reasonable solution using the following procedure. Consider the initial value problem; 

$\mathbf{y}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{+}\mathbf{2}\mathbf{y}\mathbf{\text{'}}\mathbf{+}\mathbf{5}\mathbf{y}\mathbf{=}\mathbf{g}\left(\mathbf{t}\right)\mathbf{;}\mathbf{y}\left(\mathbf{0}\right)\mathbf{=}\mathbf{0}\mathbf{,}\mathbf{y}\mathbf{\text{'}}\left(\mathbf{0}\right)\mathbf{=}\mathbf{0}$

Where,

$\mathbf{g}\left(\mathbf{t}\right)\mathbf{=}\left\{\begin{array}{l}\mathbf{10}\mathbf{,}\mathbf{if}\mathbf{0}\le \mathbf{t}\le \frac{3\pi }{2}\\ \mathbf{0}\mathbf{,}\mathbf{if}\mathbf{t}\mathbf{>}\frac{3\pi }{2}\end{array}\right\}$

1. Find a solution to the initial value problem for $\mathbf{0}\le \mathbf{t}\le \frac{3\pi }{2}$ .
2. Find a general solution for  $\mathbf{t}\mathbf{>}\frac{3\pi }{2}$. 
3. Now choose the constants in the general solution from part (b) so that the solution from part (a) and the solution from part (b) agree, together with their first derivatives, at $\mathbf{t}\mathbf{=}\frac{3\pi }{2}$ . This gives us a continuously differentiable function that satisfies the differential equation except at   $\mathbf{t}\mathbf{=}\frac{3\pi }{2}$.

Forced Vibrations. As discussed in Section 4.1, a vibrating spring with damping that is under external force can be modeled by 

 $\mathit{m}\mathit{y}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{+}\mathit{b}\mathit{y}\mathbf{\text{'}}\mathbf{+}\mathit{k}\mathit{y}\mathbf{=}\mathit{g}\left(\mathit{t}\right)$, 

Where m > 0 is the mass of the spring system, b > 0 is the damping constant, k > 0 is the spring constant, g(t) is the force on the system at time t, and y(t) is the displacement from the equilibrium of the spring system at time t. Assume ${\mathit{b}}^{\mathbf{2}}\mathbf{<}\mathbf{4}\mathit{m}\mathit{k}$.

1. Determine the form of the equation of motion for the spring system when $\mathit{g}\left(\mathit{t}\right)\mathbf{=}\mathit{s}\mathit{i}\mathit{n}\mathit{\beta }\mathit{t}$ by finding a general solution to equation (15). 
2. Discuss the long-term behavior of this system.

[Hint: Consider what happens to the general solution obtained in part (a) as $\mathit{t}\mathbf{\to }\mathbf{+}\mathbf{\infty }$.]

A mass–spring system is driven by a sinusoidal external force $\mathbf{g}\left(\mathbf{t}\right)\mathbf{=}\mathbf{5}\mathbf{sint}$. The mass equals 1, the spring constant equals 3, and the damping coefficient equals 4. If the mass is initially located at  $\mathbf{y}\left(\mathbf{0}\right)\mathbf{=}\frac{1}{2}$and at rest, i.e., $\mathbf{y}\mathbf{\text{'}}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}\mathbf{0}$, find its equation of motion.

A mass-spring system is driven by an external force $\mathbf{g}\left(\mathbf{t}\right)\mathbf{=}\mathbf{2}\mathbf{sin}\mathbf{3}\mathbf{t}\mathbf{+}\mathbf{10}\mathbf{cos}\mathbf{3}\mathbf{t}$ . The mass equals 1, the spring constant equals 5, and the damping coefficient equals 2. If the mass is initially located at  $\mathbf{y}\left(\mathbf{0}\right)\mathbf{=}\mathbf{-}\mathbf{1}$ , with an initial velocity  $\mathbf{y}\mathbf{\text{'}}\left(\mathbf{0}\right)\mathbf{=}\mathbf{5}$ , find its equation of motion.

Speed Bumps. Often bumps like the one depicted in Figure 4.11 are built into roads to discourage speeding. The figure suggests that a crude model of the vertical motion y(t) of a car encountering the speed bump with the speed V is given by

  $\mathbf{y}\left(\mathbf{t}\right)\mathbf{=}\mathbf{0}$ for $\mathbf{t}\le \frac{-L}{2V}$

$\mathbf{my}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{+}\mathbf{ky}\mathbf{=}\mathbf{\left\{}\begin{array}{l}{\mathbf{F}}_{\mathbf{0}}\mathbf{cos}\left(\frac{\mathrm{\pi Vt}}{L}\right)\mathbf{,}\mathbf{for}\mathbf{\left|}\mathbf{t}\mathbf{\right|}\mathbf{<}\frac{\mathbf{L}}{\mathbf{2}\mathbf{V}}\\ \mathbf{0}\mathbf{,}\mathbf{for}\mathbf{t}\mathbf{\ge }\frac{\mathbf{L}}{\mathbf{2}\mathbf{V}}\end{array}\mathbf{\right\}}$

(The absence of a damping term indicates that the car’s shock absorbers are not functioning.) 

1. Taking  $\mathbf{m}\mathbf{=}\mathbf{k}\mathbf{=}\mathbf{1}\mathbf{,}\mathbf{L}\mathbf{=}\mathbf{\pi }$, and ${\mathbf{F}}_0\mathbf{=}\mathbf{1}$   in appropriate units, solve this initial value problem. Thereby showing that the formula for the oscillatory motion after the car has traversed the speed bump is  $\mathbf{y}\left(\mathbf{t}\right)=\mathbf{Asint}$, where the constant A depends on the speed V. 
2. Plot the amplitude |A| of the solution y(t) found in part (a) versus the car’s speed V. From the graph, estimate the speed that produces the most violent shaking of the vehicle.

Show that the boundary value problem $\mathbf{y}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{+}{\mathbf{\lambda }}^2\mathbf{y}\mathbf{=}\mathbf{sint}\mathbf{;}\mathbf{y}\left(\mathbf{0}\right)\mathbf{=}\mathbf{0}\mathbf{,}\mathbf{y}\left(\mathbf{\pi }\right)\mathbf{=}\mathbf{1}$has a solution if and only if  $\mathbf{\lambda }\ne \mathbf{\pm }\mathbf{1}\mathbf{,}\mathbf{\pm }\mathbf{2}\mathbf{,}\mathbf{\pm }\mathbf{3}\mathbf{,}......$

Find the solution(s) to $\mathbf{y}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{+}\mathbf{9}\mathbf{y}\mathbf{=}\mathbf{27}\mathbf{cos}\left(\mathbf{6}\mathbf{t}\right)$

(If it exists) satisfying the boundary conditions.

1. $\mathbf{y}\left(\mathbf{0}\right)\mathbf{=}\mathbf{-}\mathbf{1}\mathbf{,}\mathbf{y}\mathbf{\left(}\frac{\mathbf{\pi }}{\mathbf{6}}\mathbf{\right)}\mathbf{=}\mathbf{3}$
2. $\mathbf{y}\left(\mathbf{0}\right)\mathbf{=}\mathbf{-}\mathbf{1}\mathbf{,}\mathbf{y}\mathbf{\left(}\frac{\mathbf{\pi }}{\mathbf{3}}\mathbf{\right)}\mathbf{=}\mathbf{5}$
3. $\mathbf{y}\left(\mathbf{0}\right)\mathbf{=}\mathbf{-}\mathbf{1}\mathbf{,}\mathbf{y}\mathbf{\left(}\frac{\mathbf{\pi }}{\mathbf{3}}\mathbf{\right)}\mathbf{=}\mathbf{-}\mathbf{1}$
   

All that is known concerning a mysterious second-order constant-coefficient differential equation $\mathbf{y}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{+}\mathbf{py}\mathbf{\text{'}}\mathbf{+}\mathbf{qy}\mathbf{=}\mathbf{g}\left(\mathbf{t}\right)$ is that  ${\mathbf{t}}^2\mathbf{+}\mathbf{1}\mathbf{+}{\mathbf{e}}^t\mathbf{cost}\mathbf{,}{\mathbf{t}}^2\mathbf{+}\mathbf{1}\mathbf{+}{\mathbf{e}}^t\mathbf{sint}$ and ${\mathbf{t}}^2\mathbf{+}\mathbf{1}\mathbf{+}{\mathbf{e}}^t\mathbf{cost}\mathbf{+}{\mathbf{e}}^t\mathbf{sint}$  are solutions. 

(b) Find a suitable choice of p, q, and g(t) that enables these solutions.

In Problems 1–8, find a general solution to the differential equation using the method of variation of parameters.

$\mathbf{y}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{+}\mathbf{4}\mathbf{y}\mathbf{=}\mathbf{tan}\mathbf{2}\mathbf{t}$

In Problems 1–8, find a general solution to the differential equation using the method of variation of parameters. $\mathbf{y}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{+}\mathbf{y}\mathbf{=}\mathbf{sect}$

In Problems 1–8, find a general solution to the differential equation using the method of variation of parameters.

$\mathbf{3}\mathbf{.}\mathit{y}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{-}\mathbf{2}\mathit{y}\mathbf{\text{'}}\mathbf{+}\mathit{y}\mathbf{=}{\mathit{t}}^{\mathbf{-}\mathbf{1}}{\mathit{e}}^{\mathbf{t}}$

In Problems 1–8, find a general solution to the differential equation using the method of variation of parameters.

$\mathit{y}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{+}\mathbf{2}\mathit{y}\mathbf{\text{'}}\mathbf{+}\mathit{y}\mathbf{=}{\mathit{e}}^{\mathbf{-}\mathbf{t}}$

In Problems 1–8, find a general solution to the differential equation using the method of variation of parameters.

$\mathit{y}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{\left(}\mathit{\theta }\mathbf{\right)}\mathbf{+}\mathbf{16}\mathit{y}\mathbf{\left(}\mathit{\theta }\mathbf{\right)}\mathbf{=}\mathit{s}\mathit{e}\mathit{c}\mathbf{4}\mathit{\theta }$

In Problems 1–8, find a general solution to the differential equation using the method of variation of parameters. $\mathit{y}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{+}\mathbf{9}\mathit{y}\mathbf{=}\mathit{s}\mathit{e}{\mathit{c}}^{\mathbf{2}}\mathbf{\left(}\mathbf{3}\mathbf{t}\mathbf{\right)}$

In Problems 1–8, find a general solution to the differential equation using the method of variation of parameters.

$\mathit{y}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{+}\mathbf{4}\mathit{y}\mathbf{\text{'}}\mathbf{+}\mathbf{4}\mathit{y}\mathbf{=}{\mathit{e}}^{\mathbf{-}\mathbf{2}\mathbf{t}}\mathit{l}\mathit{n}\mathit{t}$

In Problems 9 and 10, find a particular solution first by undetermined coefficients, and then by variation of parameters. Which method was quicker?

$\mathit{y}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{-}\mathit{y}\mathbf{=}\mathbf{2}\mathit{t}\mathbf{+}\mathbf{4}$

In Problems 9 and 10, find a particular solution first by undetermined coefficients, and then by variation of parameters. Which method was quicker?

$\mathbf{10}\mathbf{.}\mathbf{2}\mathbf{x}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{\left(}\mathbf{t}\mathbf{\right)}\mathbf{-}\mathbf{2}\mathbf{x}\mathbf{\text{'}}\mathbf{\left(}\mathbf{t}\mathbf{\right)}\mathbf{-}\mathbf{4}\mathbf{x}\mathbf{\left(}\mathbf{t}\mathbf{\right)}\mathbf{=}\mathbf{2}{\mathit{e}}^{\mathbf{2}\mathbf{t}}$

In Problems 11–18, find a general solution to the differential equation.

$\mathbf{y}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{+}\mathbf{y}\mathbf{=}\mathbf{tant}\mathbf{+}{\mathbf{e}}^{\mathbf{3}\mathbf{t}}\mathbf{-}\mathbf{1}$

In Problems 11–18, find a general solution to the differential equation.

$\mathbf{12}\mathbf{.}\mathbf{y}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{+}\mathbf{y}\mathbf{=}{\mathbf{tan}}^{\mathbf{2}}\mathbf{t}$

In Problems 11–18, find a general solution to the differential equation.

$\mathbf{v}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{+}\mathbf{4}\mathbf{v}\mathbf{=}{\mathbf{sec}}^{\mathbf{4}}\mathbf{\left(}\mathbf{2}\mathbf{t}\mathbf{\right)}$

In Problems 11–18, find a general solution to the differential equation

$\mathbf{y}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{\left(}\mathbf{\theta }\mathbf{\right)}\mathbf{+}\mathbf{y}\mathbf{\left(}\mathbf{\theta }\mathbf{\right)}\mathbf{=}\mathbf{se}{\mathbf{c}}^{\mathbf{3}}\mathbf{\theta }$.

In Problems 11–18, find a general solution to the differential equation.

$\mathbf{y}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{+}\mathbf{y}\mathbf{=}\mathbf{3}\mathbf{sect}\mathbf{-}{\mathbf{t}}^{\mathbf{2}}\mathbf{+}\mathbf{1}$

In Problems 11–18, find a general solution to the differential equation.

$\mathbf{y}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{+}\mathbf{5}\mathbf{y}\mathbf{\text{'}}\mathbf{+}\mathbf{6}\mathbf{y}\mathbf{=}\mathbf{18}{\mathbf{t}}^{\mathbf{2}}$

In Problems 11–18, find a general solution to the differential equation.

$\frac{\mathbf{1}}{\mathbf{2}}\mathbf{y}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{+}\mathbf{2}\mathbf{y}\mathbf{=}\mathbf{tan}\mathbf{2}\mathbf{t}\mathbf{-}\frac{\mathbf{1}}{\mathbf{2}}{\mathbf{e}}^{\mathbf{t}}$

In Problems 11–18, find a general solution to the differential equation.

$\mathbf{y}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{-}\mathbf{6}\mathbf{y}\mathbf{\text{'}}\mathbf{+}\mathbf{9}\mathbf{y}\mathbf{=}{\mathbf{t}}^{\mathbf{-}\mathbf{3}}{\mathbf{e}}^{\mathbf{3}\mathbf{t}}$

Express the solution to the initial value problem $\mathbf{y}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{-}\mathbf{y}\mathbf{=}\frac{\mathbf{1}}{\mathbf{t}}\mathbf{,}\mathbf{y}\mathbf{\left(}\mathbf{1}\mathbf{\right)}\mathbf{=}\mathbf{0}\mathbf{,}\mathbf{y}\mathbf{\text{'}}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}\mathbf{-}\mathbf{2}$, using definite integrals. Using numerical integration (Appendix C) to approximate the integrals, find an approximation for y(2) to two decimal places.

Use the method of variation of parameters to show that $\mathbf{y}\mathbf{\left(}\mathbf{t}\mathbf{\right)}\mathbf{=}{\mathbf{c}}_{\mathbf{1}}\mathbf{cost}\mathbf{+}{\mathbf{c}}_{\mathbf{2}}\mathbf{sint}\mathbf{+}\underset{\mathbf{0}}{\overset{\mathbf{t}}{\int }}\mathbf{f}\mathbf{\left(}\mathbf{s}\mathbf{\right)}\mathbf{sin}\mathbf{(}\mathbf{t}\mathbf{-}\mathbf{s}\mathbf{)}\mathbf{ds}$ is a general solution to the differential

equation $\mathbf{y}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{+}\mathbf{y}\mathbf{=}\mathbf{f}\mathbf{\left(}\mathbf{t}\mathbf{\right)}$, where f(t) is a continuous function on $\mathbf{(}\mathbf{-}\mathbf{\infty }\mathbf{,}\mathbf{\infty }\mathbf{)}$.[Hint: Use the trigonometric identity $\mathbf{sin}\mathbf{(}\mathbf{t}\mathbf{-}\mathbf{s}\mathbf{)}\mathbf{=}\mathbf{sintcoss}\mathbf{-}\mathbf{sinscost}$.]

Suppose y satisfies the equation $\mathbf{y}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{+}\mathbf{10}\mathbf{y}\mathbf{\text{'}}\mathbf{+}\mathbf{25}\mathbf{y}\mathbf{=}{\mathbf{e}}^{{\mathbf{t}}^{\mathbf{3}}}$ subject to $\mathbf{y}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}\mathbf{1}\mathbf{andy}\mathbf{\text{'}}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}\mathbf{-}\mathbf{5}$ Estimate y(0.2) to within $\mathbf{\pm }\mathbf{0}\mathbf{.}\mathbf{0001}$ by numerically approximating the integrals in the variation of parameters formula.

In Problems 22 through 25, use a variation of parameters to find a general solution to the differential equation given that the functions and are linearly independent solutions to the corresponding homogeneous equation for t> 0.${\mathbf{t}}^2\mathbf{y}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{-}\mathbf{4}\mathbf{ty}\mathbf{\text{'}}\mathbf{+}\mathbf{6}\mathbf{y}\mathbf{=}{\mathbf{t}}^3\mathbf{+}\mathbf{1}\mathbf{;}{\mathbf{y}}_1\mathbf{=}{\mathbf{t}}^2\mathbf{,}{\mathbf{y}}_2\mathbf{=}{\mathbf{t}}^3$

In Problems 22 through 25, use a variation of parameters to find a general solution to the differential equation given that the functions ${\mathbf{y}}_{\mathbf{1}}$ and ${\mathbf{y}}_2$are linearly independent solutions to the corresponding homogeneous equation for t> 0.

$\mathbf{ty}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{-}\mathbf{(}\mathbf{t}\mathbf{+}\mathbf{1}\mathbf{)}\mathbf{y}\mathbf{\text{'}}\mathbf{+}\mathbf{y}\mathbf{=}{\mathbf{t}}^2\mathbf{;}{\mathbf{y}}_1\mathbf{=}{\mathbf{e}}^t\mathbf{,}{\mathbf{y}}_2\mathbf{=}\mathbf{t}\mathbf{+}\mathbf{1}$

In Problems 22 through 25, use variation of parameters to find a general solution to the differential equation given that the functions ${\mathbf{y}}_{\mathbf{1}}$ and ${\mathbf{y}}_{\mathbf{2}}$ are linearly independent solutions to the corresponding homogeneous equation for t> 0. $\mathbf{ty}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{+}\mathbf{(}\mathbf{1}\mathbf{-}\mathbf{2}\mathbf{t}\mathbf{)}\mathbf{y}\mathbf{\text{'}}\mathbf{+}\mathbf{(}\mathbf{t}\mathbf{-}\mathbf{1}\mathbf{)}\mathbf{y}\mathbf{=}\mathbf{t}{\mathbf{e}}^t\mathbf{;}{\mathbf{y}}_1\mathbf{=}{\mathbf{e}}^t\mathbf{,}{\mathbf{y}}_2\mathbf{=}{\mathbf{e}}^t\mathbf{lnt}$

In Problems 22 through 25, use variation of parameters to find a general solution to the differential equation given that the functions ${\mathbf{y}}_{\mathbf{1}}$ and ${\mathbf{y}}_{\mathbf{2}}$ are linearly independent solutions to the corresponding homogeneous equation for t> 0.

$\mathbf{ty}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{+}\mathbf{(}\mathbf{5}\mathbf{t}\mathbf{-}\mathbf{1}\mathbf{)}\mathbf{y}\mathbf{\text{'}}\mathbf{-}\mathbf{5}\mathbf{y}\mathbf{=}{\mathbf{t}}^2{\mathbf{e}}^{-5t}\mathbf{;}{\mathbf{y}}_1\mathbf{=}\mathbf{5}\mathbf{t}\mathbf{-}\mathbf{1}\mathbf{,}{\mathbf{y}}_2\mathbf{=}{\mathbf{e}}^{-5t}$

In Problems 1 through 4, use Theorem 5 to discuss the existence and uniqueness of a solution to the differential equation that satisfies the initial conditions $\mathbf{y}\mathbf{\left(}\mathbf{1}\mathbf{\right)}\mathbf{=}{\mathbf{Y}}_{\mathbf{o}}\mathbf{,}\mathbf{y}\mathbf{\text{'}}\mathbf{\left(}\mathbf{1}\mathbf{\right)}\mathbf{=}{\mathbf{Y}}_{\mathbf{1}}$, where ${\mathbf{Y}}_{\mathbf{o}}$ and ${\mathbf{Y}}_{\mathbf{1}}$ are real constants $\mathbf{(}\mathbf{1}\mathbf{+}{\mathbf{t}}^{\mathbf{2}}\mathbf{)}\mathbf{y}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{+}\mathbf{ty}\mathbf{\text{'}}\mathbf{-}\mathbf{y}\mathbf{=}\mathbf{tant}$.

In Problems 1 through 4, use Theorem 5 to discuss the existence and uniqueness of a solution to the differential equation that satisfies the initial conditions $\mathbf{y}\mathbf{\left(}\mathbf{1}\mathbf{\right)}\mathbf{=}{\mathbf{Y}}_{\mathbf{o}}\mathbf{,}\mathbf{y}\mathbf{\text{'}}\mathbf{\left(}\mathbf{1}\mathbf{\right)}\mathbf{=}{\mathbf{Y}}_{\mathbf{1}}$, where ${\mathbf{Y}}_{\mathbf{o}}$ and ${\mathbf{Y}}_{\mathbf{1}}$ are real constants.

$\mathbf{t}\mathbf{(}\mathbf{t}\mathbf{-}\mathbf{3}\mathbf{)}\mathbf{y}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{+}\mathbf{2}\mathbf{ty}\mathbf{\text{'}}\mathbf{-}\mathbf{y}\mathbf{=}{\mathbf{t}}^{\mathbf{2}}$