Chapter 5.RP, Problem 1RP

In Problems 1-4, find a general solution $x\left(t\right)$, $y\left(t\right)$ for the given system.

$\begin{array}{rll}{x}^{\prime }+y^{″}+y& =& 0,\\ x^{″}+y^{\prime }& =& 0\end{array}$

To determine

The general solution $x\left(t\right)$ and $y\left(t\right)$ for the systems of differential equations.

Answer to Problem 1RP

Solution:

The general solution of the system is $x\left(t\right)=-\left(\frac{C_1}{3}\right)t^3-\left(\frac{C_2}{2}\right)t^2-\left(C_3+2C_1\right)t+C_4$ and $y\left(t\right)=C_1{t}^2+C_{2}t+C_3$.

Here, $C_1,C_2$ and $C_3$ are arbitrary constants.

Explanation of Solution

Given:

System of differential equations is,

$\begin{array}{rll}{x}^{\prime }+y^{″}+y& =& 0,\\ x^{″}+y^{\prime }& =& 0\end{array}$

Approach:

Use elimination method for solving algebraic systems.

Step1: Add or subtract the given equations to get an equation in single variable.

Step2: Solve the single variable equation.

Step3: substitute the value of variable obtained in Step2 in other equations to get the values of other unknowns.

Calculation:

Write the system using the operator notation.

$\begin{array}{rll}{x}^{\prime }+y^{″}+y& =& 0\\ Dx+D^{2}y+y& =& 0\end{array}$

$Dx+\left(D^2+1\right)y=0 (1)$

$x^{″}+y^{\prime }=0$

$D^{2}x+Dy=0 (2)$

Multiply Equation $(1)$ with $D$.

$D^{2}x+\left(D^2+1\right)Dy=0 (3)$

Subtract Equation $(2)$ from Equation $(3)$.

$\begin{array}{l}{D}^{2}x+\left(D^2+1\right)Dy-\left(D^{2}x+Dy\right)=0\\ D^{3}y=0 (4)\end{array}$

The auxiliary equation of Equation $(4)$ is,

$r^3=0 (5)$

Roots of Equation $(5)$ are $r=0,0,0$.

So, the general solution of Equation $(4)$ is $y\left(t\right)=C_1{t}^2+C_{2}t+C_3$.

Here, $C_1,C_2$ and $C_3$ are arbitrary constants.

$\begin{array}{l}Dy=2C_{1}t+C_2\\ D^{2}y=2C_1\end{array}$

From Equation $(1)$,

$Dx=-\left(D^2+1\right)y (6)$

Substitute $C_1{t}^2+C_{2}t+C_3$ for $y$ and $2C_1$ for $D^{2}y$ in Equation $(6)$.

$\begin{array}{rll}Dx& =& -\left(D^2+1\right)y\\ & =& -\left(C_1{t}^2+C_{2}t+C_3\right)-2C_1\\ & =& -2C_1-C_1{t}^2-C_{2}t-C_3 (7)\end{array}$

Integrating both sides with respect to $t$ in Equation $(7)$.

$\begin{array}{rll}x\left(t\right)& =& {\displaystyle \int \left(-2C_1-C_1{t}^2-C_{2}t-C_3\right)}dt\\ & =& -\left(\frac{C_1}{3}\right)t^3-\left(\frac{C_2}{2}\right)t^2-\left(C_3+2C_1\right)t+C_4\end{array}$

Therefore, the general solution of the system is $x\left(t\right)=-\left(\frac{C_1}{3}\right)t^3-\left(\frac{C_2}{2}\right)t^2-\left(C_3+2C_1\right)t+C_4$ and $y\left(t\right)=C_1{t}^2+C_{2}t+C_3$.

Here, $C_1,C_2$ and $C_3$ are arbitrary constants.

Conclusion:

Hence, the general solution of the system is $x\left(t\right)=-\left(\frac{C_1}{3}\right)t^3-\left(\frac{C_2}{2}\right)t^2-\left(C_3+2C_1\right)t+C_4$ and $y\left(t\right)=C_1{t}^2+C_{2}t+C_3$.

Here, $C_1,C_2$ and $C_3$ are arbitrary constants.