Question

In Problems 1–4 the given family of functions is the general solution of the differential equation on the indicated interval. Find a member of the family that is a solution of the initial-value problem.

$$\begin{gathered} \mathbf{y}\mathbf{=}{\mathbf{c}}_{\mathbf{1}}\mathbf{+}{\mathbf{c}}_{\mathbf{2}}\mathbf{}\mathbf{cos}\mathbf{}\mathbf{x}\mathbf{}\mathbf{+}\mathbf{}{\mathbf{c}}_{\mathbf{3}}\mathbf{}\mathbf{sin}\mathbf{}\mathbf{x}\mathbf{,}\mathbf{}\mathbf{(}\mathbf{-}\mathbf{\infty }\mathbf{,}\mathbf{\infty }\mathbf{)} \\ \mathbf{y}\mathbf{\text{'}\text{'}}\mathbf{+}\mathbf{y}\mathbf{\text{'}}\mathbf{=}\mathbf{0}\mathbf{,}\mathbf{}\mathbf{y}\mathbf{\left(}\mathbf{\pi }\mathbf{\right)}\mathbf{=}\mathbf{0}\mathbf{,}\mathbf{}\mathbf{y}\mathbf{\text{'}}\mathbf{\left(}\mathbf{\pi }\mathbf{\right)}\mathbf{=}\mathbf{2}\mathbf{,}\mathbf{}\mathbf{y}\mathbf{\text{'}\text{'}}\mathbf{\left(}\mathbf{\pi }\mathbf{\right)}\mathbf{=}\mathbf{-}\mathbf{1} \end{gathered}$$

Short answer

$\mathrm{y}=-1-\mathrm{cosx}-2\sin \mathrm{x}$

Step-by-step solution

Find the relation between first two constants of the general solution

The initial conditions are:

$$\begin{gathered} \mathrm{y}\left(\mathrm{\pi }\right)=0 \\ \mathrm{y}\text{'}\left(\mathrm{\pi }\right)=2 \\ \mathrm{y}\text{'}\text{'}\left(\mathrm{\pi }\right)=-1 \end{gathered}$$

The point lying on the general solution $\mathrm{y}={\mathrm{c}}_1+{\mathrm{c}}_2\cos \mathrm{x}+{\mathrm{c}}_3\sin \mathrm{x}$is $\left(\mathrm{x},\mathrm{y}\right)=\left(\mathrm{\pi },0\right)$.

Substituting the value in the general solution,

$$\begin{gathered} 0=c_1+c_2\cos \left(\mathrm{\pi }\right)+c_3\sin \left(\mathrm{\pi }\right) \\ 0=c_1+c_2\times \left(-1\right)+c_3\times \left(0\right) \\ {c}_1-c_2=0 \end{gathered}$$

Find the value of the third constant of the general solution

Determining the first derivative of the general solution of the differential equation,

$\mathrm{y}\text{'}=-{\mathrm{c}}_2\sin \mathrm{x}+{\mathrm{c}}_3\cos \mathrm{x}$

Substituting the value of point $\left(\mathrm{x},\mathrm{y}\text{'}\right)=\left(\mathrm{\pi },2\right)$in the first derivative,

$$\begin{gathered} 2=-c_2\sin \left(\mathrm{\pi }\right)+c_3\cos \left(\mathrm{\pi }\right) \\ 2=-c_2\times \left(0\right)+c_3\times \left(-1\right) \\ 2=0-c_3 \\ {c}_3=-2 \end{gathered}$$

Find the value of the second constant of the general solution

Determining the second derivative of the general solution of the differential equation,

$\mathrm{y}\text{'}=-{\mathrm{c}}_2\cos \mathrm{x}+{\mathrm{c}}_3\sin \mathrm{x}$

Substituting the value of point $\left(\mathrm{x},\mathrm{y}\text{'}\text{'}\right)=\left(\mathrm{\pi },-1\right)$in the first derivative,

$$\begin{gathered} -1=-{\mathrm{c}}_2\cos \left(\mathrm{\pi }\right)-{\mathrm{c}}_3\sin \left(\mathrm{\pi }\right) \\ -1=-{\mathrm{c}}_2\times \left(-1\right)-{\mathrm{c}}_3\times \left(0\right) \\ {\mathrm{c}}_2=-1 \end{gathered}$$

Find the value of the first constant of the general solution

Using the relation between the first and second constants of the general solution,

${\mathrm{c}}_1-{\mathrm{c}}_2=0$

Substituting ${\mathrm{c}}_2=-1$

$$\begin{gathered} {\mathrm{c}}_1-\left(-1\right)=0 \\ {\mathrm{c}}_1=-1 \end{gathered}$$

Find the general solution of the differential equation

Substituting the values of the constants in the general solution,

$$\begin{gathered} \mathrm{y}=-1+\left(-1\right)\cos \mathrm{x}+\left(-2\right)\sin \mathrm{x} \\ \mathrm{y}=-1-\cos \mathrm{x}-2\sin \mathrm{x} \end{gathered}$$

localid="1668489021191" $\mathbf{Hence}\mathbf{}\mathbf{the}\mathbf{}\mathbf{solution}\mathbf{}\mathbf{is}\mathbf{}\mathbf{}\mathbf{y}\mathbf{=}\mathbf{-}\mathbf{1}\mathbf{-}\mathit{c}\mathit{o}\mathit{s}\mathbf{}\mathbf{x}\mathbf{-}\mathbf{2}\mathit{s}\mathit{i}\mathit{n}\mathbf{}\mathbf{x}$