2.1.88 Problem 87

Internal problem ID [8800]
Book : Own collection of miscellaneous problems
Section : section 1.0
Problem number : 87
Date solved : Wednesday, March 05, 2025 at 06:51:23 AM
CAS classification : [[_2nd_order, _missing_x]]

Solve

$$\begin{array}{rl}{y}^{\prime \prime }+y^{\prime }+y& =0\end{array}$$

With initial conditions

$$\begin{array}{rl}y\left(0\right)& =0\end{array}$$

Solved as second order linear constant coeff ode

Time used: 0.099 (sec)

This is second order with constant coefficients homogeneous ODE. In standard form the ODE is

$$A{y}^{″}(x)+B{y}^{\prime }(x)+Cy(x)=0$$

Where in the above $A=1,B=1,C=1$. Let the solution be $y=e^{\lambda x}$. Substituting this into the ODE gives

$$\begin{array}{}\text{(1)}& {\lambda }^2{\mathrm{e}}^{x\lambda }+\lambda {\mathrm{e}}^{x\lambda }+{\mathrm{e}}^{x\lambda }=0\end{array}$$

Since exponential function is never zero, then dividing Eq(2) throughout by $e^{\lambda x}$ gives

$$\begin{array}{}\text{(2)}& {\lambda }^2+\lambda +1=0\end{array}$$

Equation (2) is the characteristic equation of the ODE. Its roots determine the general solution form.Using the quadratic formula

$${\lambda }_{1,2}=\frac{-B}{2A}\pm \frac{1}{2A}\sqrt{B^2-4AC}$$

Substituting $A=1,B=1,C=1$ into the above gives

$$\begin{array}{rl}{\lambda }_{1,2}& =\frac{-1}{(2)\left(1\right)}\pm \frac{1}{(2)\left(1\right)}\sqrt{1^2-(4)\left(1\right)\left(1\right)}\\ & =-\frac{1}{2}\pm \frac{i\sqrt{3}}{2}\end{array}$$

Hence

$$\begin{array}{rl}{\lambda }_1& =-\frac{1}{2}+\frac{i\sqrt{3}}{2}\\ {\lambda }_2& =-\frac{1}{2}-\frac{i\sqrt{3}}{2}\end{array}$$

Which simplifies to

$$\begin{array}{rl}{\lambda }_1& =-\frac{1}{2}+\frac{i\sqrt{3}}{2}\\ {\lambda }_2& =-\frac{1}{2}-\frac{i\sqrt{3}}{2}\end{array}$$

Since roots are complex conjugate of each others, then let the roots be

$${\lambda }_{1,2}=\alpha \pm i\beta$$

Where $\alpha =-\frac{1}{2}$ and $\beta =\frac{\sqrt{3}}{2}$. Therefore the final solution, when using Euler relation, can be written as

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

Which becomes

$$y=e^{-\frac{x}{2}}(c_1\cos \left(\frac{\sqrt{3}x}{2}\right)+c_2\sin \left(\frac{\sqrt{3}x}{2}\right))$$

Will add steps showing solving for IC soon.

Summary of solutions found

$$\begin{array}{rl}y& ={\mathrm{e}}^{-\frac{x}{2}}{c}_2\sin \left(\frac{\sqrt{3}x}{2}\right)\end{array}$$

Figure 2.172: Slope field $y^{\prime \prime }+y^{\prime }+y=0$

Solved as second order ode using Kovacic algorithm

Time used: 0.243 (sec)

Writing the ode as

$$\begin{array}{}\text{(1)}& y^{\prime \prime }+y^{\prime }+y& =0\text{(2)}& A{y}^{\prime \prime }+B{y}^{\prime }+Cy& =0\end{array}$$

Comparing (1) and (2) shows that

$$\begin{array}{rl}A& =1\\ \text{(3)}& B& =1C& =1\end{array}$$

Applying the Liouville transformation on the dependent variable gives

$$\begin{array}{rl}z(x)& =y{e}^{\int \frac{B}{2A}dx}\end{array}$$

Then (2) becomes

$$\begin{array}{}\text{(4)}& z^{″}(x)=rz(x)\end{array}$$

Where $r$ is given by

$$\begin{array}{rl}\text{(5)}& r& =\frac{s}{t}& =\frac{2A{B}^{\prime }-2B{A}^{\prime }+B^2-4AC}{4A^2}\end{array}$$

Substituting the values of $A,B,C$ from (3) in the above and simplifying gives

$$\begin{array}{}\text{(6)}& r& =\frac{-3}{4}\end{array}$$

Comparing the above to (5) shows that

$$\begin{array}{rl}s& =-3\\ t& =4\end{array}$$

Therefore eq. (4) becomes

$$\begin{array}{}\text{(7)}& z^{″}(x)& =-\frac{3z\left(x\right)}{4}\end{array}$$

Equation (7) is now solved. After finding $z(x)$ then $y$ is found using the inverse transformation

$$\begin{array}{rl}y& =z\left(x\right)e^{-\int \frac{B}{2A}dx}\end{array}$$

The first step is to determine the case of Kovacic algorithm this ode belongs to. There are 3 cases depending on the order of poles of $r$ and the order of $r$ at $\mathrm{\infty }$. The following table summarizes these cases.

Case	Allowed pole order for $r$	Allowed value for $\mathcal{O}(\mathrm{\infty })$
1	$\{0,1,2,4,6,8,\cdots \}$	$\{\cdots ,-6,-4,-2,0,2,3,4,5,6,\cdots \}$
2	Need to have at least one pole that is either order $2$ or odd order greater than $2$. Any other pole order is allowed as long as the above condition is satisfied. Hence the following set of pole orders are all allowed. $\{1,2\}$,$\{1,3\}$,$\{2\}$,$\{3\}$,$\{3,4\}$,$\{1,2,5\}$.	no condition
3	$\{1,2\}$	$\{2,3,4,5,6,7,\cdots \}$

Table 2.17: Necessary conditions for each Kovacic case

The order of $r$ at $\mathrm{\infty }$ is the degree of $t$ minus the degree of $s$. Therefore

$$\begin{array}{rl}O\left(\mathrm{\infty }\right)& =\text{deg}(t)-\text{deg}(s)\\ & =0-0\\ & =0\end{array}$$

There are no poles in $r$. Therefore the set of poles $\mathrm{\Gamma }$ is empty. Since there is no odd order pole larger than $2$ and the order at $\mathrm{\infty }$ is $0$ then the necessary conditions for case one are met. Therefore

$$\begin{array}{rl}L& =[1]\end{array}$$

Since $r=-\frac{3}{4}$ is not a function of $x$, then there is no need run Kovacic algorithm to obtain a solution for transformed ode $z^{″}=rz$ as one solution is

$$z_1(x)=\cos \left(\frac{\sqrt{3}x}{2}\right)$$

Using the above, the solution for the original ode can now be found. The first solution to the original ode in $y$ is found from

$$\begin{array}{rl}{y}_1& =z_1{e}^{\int -\frac{1}{2}\frac{B}{A}dx}\\ & =z_1{e}^{-\int \frac{1}{2}\frac{1}{1}dx}\\ & =z_1{e}^{-\frac{x}{2}}\\ & =z_1\left({\mathrm{e}}^{-\frac{x}{2}}\right)\end{array}$$

Which simplifies to

$$y_1={\mathrm{e}}^{-\frac{x}{2}}\cos \left(\frac{\sqrt{3}x}{2}\right)$$

The second solution $y_2$ to the original ode is found using reduction of order

$$y_2=y_1\int \frac{e^{\int -\frac{B}{A}dx}}{y_1^2}dx$$

Substituting gives

$$\begin{array}{rl}{y}_2& =y_1\int \frac{e^{\int -\frac{1}{1}dx}}{{\left(y_1\right)}^2}dx\\ & =y_1\int \frac{e^{-x}}{{\left(y_1\right)}^2}dx\\ & =y_1\left(\frac{2\sqrt{3}\tan \left(\frac{\sqrt{3}x}{2}\right)}{3}\right)\end{array}$$

Therefore the solution is

$$\begin{array}{rl}y& =c_1{y}_1+c_2{y}_2\\ & =c_1({\mathrm{e}}^{-\frac{x}{2}}\cos \left(\frac{\sqrt{3}x}{2}\right))+c_2({\mathrm{e}}^{-\frac{x}{2}}\cos \left(\frac{\sqrt{3}x}{2}\right)\left(\frac{2\sqrt{3}\tan \left(\frac{\sqrt{3}x}{2}\right)}{3}\right))\end{array}$$

Will add steps showing solving for IC soon.

Summary of solutions found

$$\begin{array}{rl}y& =\frac{2c_2{\mathrm{e}}^{-\frac{x}{2}}\sqrt{3}\sin \left(\frac{\sqrt{3}x}{2}\right)}{3}\end{array}$$

Figure 2.173: Slope field $y^{\prime \prime }+y^{\prime }+y=0$

Solved as second order ode adjoint method

Time used: 1.312 (sec)

In normal form the ode

$$\begin{array}{}\text{(1)}& y^{\prime \prime }+y^{\prime }+y=0\end{array}$$

Becomes

$$\begin{array}{}\text{(2)}& y^{\prime \prime }+p\left(x\right)y^{\prime }+q\left(x\right)y& =r\left(x\right)\end{array}$$

Where

$$\begin{array}{rl}p\left(x\right)& =1\\ q\left(x\right)& =1\\ r\left(x\right)& =0\end{array}$$

The Lagrange adjoint ode is given by

$$\begin{array}{rl}{\xi }^{{}^{″}}-(\xi p{)}^{\prime }+\xi q& =0\\ {\xi }^{{}^{″}}-{\left(\xi \left(x\right)\right)}^{\prime }+\left(\xi \left(x\right)\right)& =0\\ {\xi }^{\prime \prime }\left(x\right)-{\xi }^{\prime }\left(x\right)+\xi \left(x\right)& =0\end{array}$$

Which is solved for $\xi (x)$. This is second order with constant coefficients homogeneous ODE. In standard form the ODE is

$$A{\xi }^{″}(x)+B{\xi }^{\prime }(x)+C\xi (x)=0$$

Where in the above $A=1,B=-1,C=1$. Let the solution be $\xi =e^{\lambda x}$. Substituting this into the ODE gives

$$\begin{array}{}\text{(1)}& {\lambda }^2{\mathrm{e}}^{x\lambda }-\lambda {\mathrm{e}}^{x\lambda }+{\mathrm{e}}^{x\lambda }=0\end{array}$$

Since exponential function is never zero, then dividing Eq(2) throughout by $e^{\lambda x}$ gives

$$\begin{array}{}\text{(2)}& {\lambda }^2-\lambda +1=0\end{array}$$

Equation (2) is the characteristic equation of the ODE. Its roots determine the general solution form.Using the quadratic formula

$${\lambda }_{1,2}=\frac{-B}{2A}\pm \frac{1}{2A}\sqrt{B^2-4AC}$$

Substituting $A=1,B=-1,C=1$ into the above gives

$$\begin{array}{rl}{\lambda }_{1,2}& =\frac{1}{(2)\left(1\right)}\pm \frac{1}{(2)\left(1\right)}\sqrt{-1^2-(4)\left(1\right)\left(1\right)}\\ & =\frac{1}{2}\pm \frac{i\sqrt{3}}{2}\end{array}$$

Hence

$$\begin{array}{rl}{\lambda }_1& =\frac{1}{2}+\frac{i\sqrt{3}}{2}\\ {\lambda }_2& =\frac{1}{2}-\frac{i\sqrt{3}}{2}\end{array}$$

Which simplifies to

$$\begin{array}{rl}{\lambda }_1& =\frac{1}{2}+\frac{i\sqrt{3}}{2}\\ {\lambda }_2& =\frac{1}{2}-\frac{i\sqrt{3}}{2}\end{array}$$

Since roots are complex conjugate of each others, then let the roots be

$${\lambda }_{1,2}=\alpha \pm i\beta$$

Where $\alpha =\frac{1}{2}$ and $\beta =\frac{\sqrt{3}}{2}$. Therefore the final solution, when using Euler relation, can be written as

$$\xi =e^{\alpha x}(c_1\cos (\beta x)+c_2\sin (\beta x))$$

Which becomes

$$\xi =e^{\frac{x}{2}}(c_1\cos \left(\frac{\sqrt{3}x}{2}\right)+c_2\sin \left(\frac{\sqrt{3}x}{2}\right))$$

Will add steps showing solving for IC soon.

The original ode now reduces to first order ode

$$\begin{array}{rl}\xi \left(x\right)y^{\prime }-y{\xi }^{\prime }\left(x\right)+\xi \left(x\right)p\left(x\right)y& =\int \xi \left(x\right)r\left(x\right)dx\\ y^{\prime }+y(p\left(x\right)-\frac{{\xi }^{\prime }\left(x\right)}{\xi \left(x\right)})& =\frac{\int \xi \left(x\right)r\left(x\right)dx}{\xi \left(x\right)}\end{array}$$

Or

$$\begin{array}{rl}{y}^{\prime }+y(1-\frac{(\frac{{\mathrm{e}}^{\frac{x}{2}}(c_1\cos \left(\frac{\sqrt{3}x}{2}\right)+c_2\sin \left(\frac{\sqrt{3}x}{2}\right))}{2}+{\mathrm{e}}^{\frac{x}{2}}(-\frac{c_1\sqrt{3}\sin \left(\frac{\sqrt{3}x}{2}\right)}{2}+\frac{c_2\sqrt{3}\cos \left(\frac{\sqrt{3}x}{2}\right)}{2})){\mathrm{e}}^{-\frac{x}{2}}}{c_1\cos \left(\frac{\sqrt{3}x}{2}\right)+c_2\sin \left(\frac{\sqrt{3}x}{2}\right)})& =0\end{array}$$

Which is now a first order ode. This is now solved for $y$. In canonical form a linear first order is

$$\begin{array}{rl}{y}^{\prime }+q(x)y& =p(x)\end{array}$$

Comparing the above to the given ode shows that

$$\begin{array}{rl}q(x)& =-\frac{(c_2\sqrt{3}-c_1)\cos \left(\frac{\sqrt{3}x}{2}\right)-\sin \left(\frac{\sqrt{3}x}{2}\right)(c_1\sqrt{3}+c_2)}{2c_1\cos \left(\frac{\sqrt{3}x}{2}\right)+2c_2\sin \left(\frac{\sqrt{3}x}{2}\right)}\\ p(x)& =0\end{array}$$

The integrating factor $\mu$ is

$$\begin{array}{rl}\mu & =e^{\int qdx}\\ & ={\mathrm{e}}^{\int -\frac{(c_2\sqrt{3}-c_1)\cos \left(\frac{\sqrt{3}x}{2}\right)-\sin \left(\frac{\sqrt{3}x}{2}\right)(c_1\sqrt{3}+c_2)}{2c_1\cos \left(\frac{\sqrt{3}x}{2}\right)+2c_2\sin \left(\frac{\sqrt{3}x}{2}\right)}dx}\\ & =\frac{\sqrt{\sec {\left(\frac{\sqrt{3}x}{2}\right)}^2}{\mathrm{e}}^{\frac{\sqrt{3}\arctan (\tan \left(\frac{\sqrt{3}x}{2}\right))}{3}}}{\tan \left(\frac{\sqrt{3}x}{2}\right)c_2+c_1}\end{array}$$

The ode becomes

$$\begin{array}{rl}\frac{\mathrm{d}}{\mathrm{d}x}\mu y& =0\\ \frac{\mathrm{d}}{\mathrm{d}x}\left(\frac{y\sqrt{\sec {\left(\frac{\sqrt{3}x}{2}\right)}^2}{\mathrm{e}}^{\frac{\sqrt{3}\arctan (\tan \left(\frac{\sqrt{3}x}{2}\right))}{3}}}{\tan \left(\frac{\sqrt{3}x}{2}\right)c_2+c_1}\right)& =0\end{array}$$

Integrating gives

$$\begin{array}{rl}\frac{y\sqrt{\sec {\left(\frac{\sqrt{3}x}{2}\right)}^2}{\mathrm{e}}^{\frac{\sqrt{3}\arctan (\tan \left(\frac{\sqrt{3}x}{2}\right))}{3}}}{\tan \left(\frac{\sqrt{3}x}{2}\right)c_2+c_1}& =\int 0dx+c_3\\ & =c_3\end{array}$$

Dividing throughout by the integrating factor $\frac{\sqrt{\sec {\left(\frac{\sqrt{3}x}{2}\right)}^2}{\mathrm{e}}^{\frac{\sqrt{3}\arctan (\tan \left(\frac{\sqrt{3}x}{2}\right))}{3}}}{\tan \left(\frac{\sqrt{3}x}{2}\right)c_2+c_1}$ gives the final solution

$$y=\frac{(\tan \left(\frac{\sqrt{3}x}{2}\right)c_2+c_1){\mathrm{e}}^{-\frac{\sqrt{3}\arctan (\tan \left(\frac{\sqrt{3}x}{2}\right))}{3}}{c}_3}{\sqrt{\sec {\left(\frac{\sqrt{3}x}{2}\right)}^2}}$$

Hence, the solution found using Lagrange adjoint equation method is

$$\begin{array}{rl}y& =\frac{(\tan \left(\frac{\sqrt{3}x}{2}\right)c_2+c_1){\mathrm{e}}^{-\frac{\sqrt{3}\arctan (\tan \left(\frac{\sqrt{3}x}{2}\right))}{3}}{c}_3}{\sqrt{\sec {\left(\frac{\sqrt{3}x}{2}\right)}^2}}\end{array}$$

The constants can be merged to give

$$y=\frac{(\tan \left(\frac{\sqrt{3}x}{2}\right)c_2+c_1){\mathrm{e}}^{-\frac{\sqrt{3}\arctan (\tan \left(\frac{\sqrt{3}x}{2}\right))}{3}}}{\sqrt{\sec {\left(\frac{\sqrt{3}x}{2}\right)}^2}}$$

Will add steps showing solving for IC soon.

Summary of solutions found

$$\begin{array}{rl}y& =\frac{\tan \left(\frac{\sqrt{3}x}{2}\right)c_2{\mathrm{e}}^{-\frac{\sqrt{3}\arctan (\tan \left(\frac{\sqrt{3}x}{2}\right))}{3}}}{\sqrt{\sec {\left(\frac{\sqrt{3}x}{2}\right)}^2}}\end{array}$$

Figure 2.174: Slope field $y^{\prime \prime }+y^{\prime }+y=0$

✓ Maple. Time used: 0.013 (sec). Leaf size: 17

ode:=diff(diff(y(x),x),x)+diff(y(x),x)+y(x) = 0; 
ic:=y(0) = 0; 
dsolve([ode,ic],y(x), singsol=all);
 

$$y=c_1{\mathrm{e}}^{-\frac{x}{2}}\sin \left(\frac{\sqrt{3}x}{2}\right)$$

Maple trace

`Methods for second order ODEs: 
--- Trying classification methods --- 
trying a quadrature 
checking if the LODE has constant coefficients 
<- constant coefficients successful`
 

Maple step by step

$$\begin{array}{lll}& & \text{Let’s solve}\\ & & [\frac{d^2}{d{x}^2}y\left(x\right)+\frac{d}{dx}y\left(x\right)+y\left(x\right)=0,y\left(0\right)=0]\\ \bullet & & \text{Highest derivative means the order of the ODE is}2\\ & & \frac{d^2}{d{x}^2}y\left(x\right)\\ \bullet & & \text{Characteristic polynomial of ODE}\\ & & r^2+r+1=0\\ \bullet & & \text{Use quadratic formula to solve for}r\\ & & r=\frac{(-1)\pm \left(\sqrt{-3}\right)}{2}\\ \bullet & & \text{Roots of the characteristic polynomial}\\ & & r=(-\frac{1}{2}-\frac{\mathrm{I}\sqrt{3}}{2},-\frac{1}{2}+\frac{\mathrm{I}\sqrt{3}}{2})\\ \bullet & & \text{1st solution of the ODE}\\ & & y_1\left(x\right)={\mathrm{e}}^{-\frac{x}{2}}\cos \left(\frac{\sqrt{3}x}{2}\right)\\ \bullet & & \text{2nd solution of the ODE}\\ & & y_2\left(x\right)={\mathrm{e}}^{-\frac{x}{2}}\sin \left(\frac{\sqrt{3}x}{2}\right)\\ \bullet & & \text{General solution of the ODE}\\ & & y\left(x\right)=\mathit{C}\mathit{1}{y}_1\left(x\right)+\mathit{C}\mathit{2}{y}_2\left(x\right)\\ \bullet & & \text{Substitute in solutions}\\ & & y\left(x\right)=\mathit{C}\mathit{1}{\mathrm{e}}^{-\frac{x}{2}}\cos \left(\frac{\sqrt{3}x}{2}\right)+\mathit{C}\mathit{2}{\mathrm{e}}^{-\frac{x}{2}}\sin \left(\frac{\sqrt{3}x}{2}\right)\end{array}$$

✓ Mathematica. Time used: 0.022 (sec). Leaf size: 26

ode=D[y[x],{x,2}]+D[y[x],x]+y[x]==0; 
ic={y[0]==0}; 
DSolve[{ode,ic},y[x],x,IncludeSingularSolutions->True]
 

$$y(x)\to c_1{e}^{-x/2}\sin \left(\frac{\sqrt{3}x}{2}\right)$$

✓ Sympy. Time used: 0.147 (sec). Leaf size: 19

from sympy import * 
x = symbols("x") 
y = Function("y") 
ode = Eq(y(x) + Derivative(y(x), x) + Derivative(y(x), (x, 2)),0) 
ics = {y(0): 0} 
dsolve(ode,func=y(x),ics=ics)
 

$$y\left(x\right)=C_1{e}^{-\frac{x}{2}}\sin \left(\frac{\sqrt{3}x}{2}\right)$$