Chapter 7.1, Problem 80E

a.

To determine

To find:

To find at least one particular solution for $y^{″}=x$

Answer to Problem 80E

The solution is $y=\frac{1}{6}{x}^3+c_{1}x+c_2$

Explanation of Solution

Given information:

The given derivative is $y^{″}=x$ .

Consider the derivative $y^{″}=x$ ,

Find the anti-derivative twice to find y,

  $\begin{array}{l}{y}^{″}=x\\ y^{\prime }=\frac{1}{2}{x}^2+c_1\\ y=\frac{1}{6}{x}^3+c_{1}x+c_2\end{array}$

Therefore, the solution is $y=\frac{1}{6}{x}^3+c_{1}x+c_2$

b.

To determine

To find:

To find at least one particular solution for $y^{″}=-x$

Answer to Problem 80E

The solution is $y=-\frac{1}{6}{x}^3+c_{1}x+c_2$

Explanation of Solution

Given information:

The given derivative is $y^{″}=-x$ .

Consider the derivative $y^{″}=-x$ ,

Find the anti-derivative twice to find y,

  $\begin{array}{l}{y}^{″}=-x\\ y^{\prime }=-\frac{1}{2}{x}^2+c_1\\ y=-\frac{1}{6}{x}^3+c_{1}x+c_2\end{array}$

Therefore, the solution is $y=-\frac{1}{6}{x}^3+c_{1}x+c_2$

c.

To determine

To find:

To find at least one particular solution for $y^{″}=-\sin x$

Answer to Problem 80E

The solution is $y=\sin x+c_{1}x+c_2$

Explanation of Solution

Given information:

The given derivative is $y^{″}=-\sin x$ .

Consider the derivative $y^{″}=-\sin x$ ,

Find the anti-derivative twice to find y,

  $\begin{array}{l}{y}^{″}=-\sin x\\ y^{\prime }=\cos x+c_1\\ y=\sin x+c_{1}x+c_2\end{array}$

Therefore, the solution is $y=\sin x+c_{1}x+c_2$

d.

To determine

To find:

To find at least one particular solution for $y^{″}=y$

Answer to Problem 80E

The solution is $y=c_1{e}^t+c_2{e}^{-t}$

Explanation of Solution

Given information:

The given derivative is $y^{″}=y$ .

Consider the derivative $y^{″}=y$ ,

A second order linear homogeneous differential equation with constant co-efficients has the form

  $a{y}^{″}+b{y}^{\prime }+cy=0$

Rewrite in the form of a second order differential equation, $y^{″}-y=0$

The auxillary equation can be written as

  $\begin{array}{l}{m}^2-1=0\\ m^2=1\\ m=\pm 1\end{array}$

Therefore, the roots are $m_1=1$ and $m_2=-1$

For two real roots $m_1\ne m_2$ , the general solution takes the form

  $y=c_1{e}^{m_{1}t}+c_2{e}^{m_{2}t}$

Hence, the general solution is

  $y=c_1{e}^t+c_2{e}^{-t}$

Therefore, the general solution is $y=c_1{e}^t+c_2{e}^{-t}$

e.

To determine

To find:

To find at least one particular solution for $y^{″}=-y$

Answer to Problem 80E

The solution is $y=c_1\cos t+c_2\sin t$

Explanation of Solution

Given information:

The given derivative is $y^{″}=-y$ .

Consider the derivative $y^{″}=-y$ ,

A second order linear homogeneous differential equation with constant co-efficients has the form

  $a{y}^{″}+b{y}^{\prime }+cy=0$

Rewrite in the form of a second order differential equation,

  $y^{″}+y=0$

The auxillary equation can be written as

  $\begin{array}{l}{m}^2+1=0\\ m^2=-1\\ m=\pm 1i\end{array}$

Therefore, the roots are $m_1=i$ and $m_2=-i$

For two complex roots $m_1\ne m_2$ , the general solution takes the form

  $y=e^{\alpha t}\left(c_1\cos \left(\beta t\right)+c_2\sin \left(\beta t\right)\right)$

Where $m_1=\alpha +i\beta$ and $m_2=\alpha -i\beta$

Hence, the general solution is

  $\begin{array}{l}y=e^0\left(c_1\cos t+c_2\sin t\right)\\ y=c_1\cos t+c_2\sin t\end{array}$

Therefore, the general solution is $y=c_1\cos t+c_2\sin t$