2.1.38 Problem 38

Internal problem ID [9109]
Book : Second order enumerated odes
Section : section 1
Problem number : 38
Date solved : Sunday, March 30, 2025 at 02:07:20 PM
CAS classification : [[_2nd_order, _missing_x]]

Solved as second order linear constant coeff ode

Time used: 0.113 (sec)

Solve

\begin{align*} y^{\prime \prime }+y&=1 \end{align*}

This is second order non-homogeneous ODE. In standard form the ODE is

\[ A y''(x) + B y'(x) + C y(x) = f(x) \]

Where \(A=1, B=0, C=1, f(x)=1\). Let the solution be

\[ y = y_h + y_p \]

Where \(y_h\) is the solution to the homogeneous ODE \( A y''(x) + B y'(x) + C y(x) = 0\), and \(y_p\) is a particular solution to the non-homogeneous ODE \(A y''(x) + B y'(x) + C y(x) = f(x)\). \(y_h\) is the solution to

\[ y^{\prime \prime }+y = 0 \]

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

\[ A y''(x) + B y'(x) + C y(x) = 0 \]

Where in the above \(A=1, B=0, C=1\). Let the solution be \(y=e^{\lambda x}\). Substituting this into the ODE gives

\[ \lambda ^{2} {\mathrm e}^{x \lambda }+{\mathrm e}^{x \lambda } = 0 \tag {1} \]

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

\[ \lambda ^{2}+1 = 0 \tag {2} \]

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

\[ \lambda _{1,2} = \frac {-B}{2 A} \pm \frac {1}{2 A} \sqrt {B^2 - 4 A C} \]

Substituting \(A=1, B=0, C=1\) into the above gives

\begin{align*} \lambda _{1,2} &= \frac {0}{(2) \left (1\right )} \pm \frac {1}{(2) \left (1\right )} \sqrt {0^2 - (4) \left (1\right )\left (1\right )}\\ &= \pm i \end{align*}

Hence

\begin{align*} \lambda _1 &= + i\\ \lambda _2 &= - i \end{align*}

Which simplifies to

\begin{align*} \lambda _1 &= i \\ \lambda _2 &= -i \\ \end{align*}

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

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

Where \(\alpha =0\) and \(\beta =1\). Therefore the final solution, when using Euler relation, can be written as

\[ y = e^{\alpha x} \left ( c_1 \cos (\beta x) + c_2 \sin (\beta x) \right ) \]

Which becomes

\[ y = e^{0}\left (c_1 \cos \left (x \right )+c_2 \sin \left (x \right )\right ) \]

Or

\[ y = c_1 \cos \left (x \right )+c_2 \sin \left (x \right ) \]

Therefore the homogeneous solution \(y_h\) is

\[ y_h = c_1 \cos \left (x \right )+c_2 \sin \left (x \right ) \]

The particular solution is now found using the method of undetermined coefficients. Looking at the RHS of the ode, which is

\[ 1 \]

Shows that the corresponding undetermined set of the basis functions (UC_set) for the trial solution is

\[ [\{1\}] \]

While the set of the basis functions for the homogeneous solution found earlier is

\[ \{\cos \left (x \right ), \sin \left (x \right )\} \]

Since there is no duplication between the basis function in the UC_set and the basis functions of the homogeneous solution, the trial solution is a linear combination of all the basis in the UC_set.

\[ y_p = A_{1} \]

The unknowns \(\{A_{1}\}\) are found by substituting the above trial solution \(y_p\) into the ODE and comparing coefficients. Substituting the trial solution into the ODE and simplifying gives

\[ A_{1} = 1 \]

Solving for the unknowns by comparing coefficients results in

\[ [A_{1} = 1] \]

Substituting the above back in the above trial solution \(y_p\), gives the particular solution

\[ y_p = 1 \]

Therefore the general solution is

\begin{align*} y &= y_h + y_p \\ &= \left (c_1 \cos \left (x \right )+c_2 \sin \left (x \right )\right ) + \left (1\right ) \\ \end{align*}

Will add steps showing solving for IC soon.

Summary of solutions found

\begin{align*} y &= 1+c_1 \cos \left (x \right )+c_2 \sin \left (x \right ) \\ \end{align*}

Figure 2.105: Slope field \(y^{\prime \prime }+y = 1\)

Solved as second order can be made integrable

Time used: 1.094 (sec)

Solve

\begin{align*} y^{\prime \prime }+y&=1 \end{align*}

Multiplying the ode by \(y^{\prime }\) gives

\[ y^{\prime } y^{\prime \prime }+y^{\prime } y-y^{\prime } = 0 \]

Integrating the above w.r.t \(x\) gives

\begin{align*} \int \left (y^{\prime } y^{\prime \prime }+y^{\prime } y-y^{\prime }\right )d x &= 0 \\ \frac {y^{2}}{2}-y+\frac {{y^{\prime }}^{2}}{2} &= c_1 \end{align*}

Which is now solved for \(y\). Solving for the derivative gives these ODE’s to solve

\begin{align*} \tag{1} y^{\prime }&=\sqrt {-y^{2}+2 y+2 c_1} \\ \tag{2} y^{\prime }&=-\sqrt {-y^{2}+2 y+2 c_1} \\ \end{align*}

Now each of the above is solved separately.

Solving Eq. (1)

Integrating gives

\begin{align*} \int \frac {1}{\sqrt {-y^{2}+2 c_1 +2 y}}d y &= dx\\ \arctan \left (\frac {y -1}{\sqrt {-y^{2}+2 c_1 +2 y}}\right )&= x +c_2 \end{align*}

Singular solutions are found by solving

\begin{align*} \sqrt {-y^{2}+2 c_1 +2 y}&= 0 \end{align*}

for \(y\). This is because we had to divide by this in the above step. This gives the following singular solution(s), which also have to satisfy the given ODE.

\begin{align*} y = 1-\sqrt {2 c_1 +1}\\ y = 1+\sqrt {2 c_1 +1} \end{align*}

Solving Eq. (2)

Integrating gives

\begin{align*} \int -\frac {1}{\sqrt {-y^{2}+2 c_1 +2 y}}d y &= dx\\ -\arctan \left (\frac {y -1}{\sqrt {-y^{2}+2 c_1 +2 y}}\right )&= x +c_3 \end{align*}

Singular solutions are found by solving

\begin{align*} -\sqrt {-y^{2}+2 c_1 +2 y}&= 0 \end{align*}

for \(y\). This is because we had to divide by this in the above step. This gives the following singular solution(s), which also have to satisfy the given ODE.

\begin{align*} y = 1-\sqrt {2 c_1 +1}\\ y = 1+\sqrt {2 c_1 +1} \end{align*}

Will add steps showing solving for IC soon.

Solving for \(y\) from the above solution(s) gives (after possible removing of solutions that do not verify)

\begin{align*} y&=1-\sqrt {2 c_1 +1}\\ y&=1+\sqrt {2 c_1 +1}\\ y&=\tan \left (x +c_2 \right ) \sqrt {\frac {2 c_1 +1}{\tan \left (x +c_2 \right )^{2}+1}}+1 \end{align*}

The solution

\[ y = 1-\sqrt {2 c_1 +1} \]

was found not to satisfy the ode or the IC. Hence it is removed. The solution

\[ y = 1+\sqrt {2 c_1 +1} \]

was found not to satisfy the ode or the IC. Hence it is removed.

Summary of solutions found

\begin{align*} y &= \tan \left (x +c_2 \right ) \sqrt {\frac {2 c_1 +1}{\tan \left (x +c_2 \right )^{2}+1}}+1 \\ \end{align*}

Figure 2.106: Slope field \(y^{\prime \prime }+y = 1\)

Solved as second order ode using Kovacic algorithm

Time used: 0.155 (sec)

Solve

\begin{align*} y^{\prime \prime }+y&=1 \end{align*}

Writing the ode as

\begin{align*} y^{\prime \prime }+y &= 0 \tag {1} \\ A y^{\prime \prime } + B y^{\prime } + C y &= 0 \tag {2} \end{align*}

Comparing (1) and (2) shows that

\begin{align*} A &= 1 \\ B &= 0\tag {3} \\ C &= 1 \end{align*}

Applying the Liouville transformation on the dependent variable gives

\begin{align*} z(x) &= y e^{\int \frac {B}{2 A} \,dx} \end{align*}

Then (2) becomes

\begin{align*} z''(x) = r z(x)\tag {4} \end{align*}

Where \(r\) is given by

\begin{align*} r &= \frac {s}{t}\tag {5} \\ &= \frac {2 A B' - 2 B A' + B^2 - 4 A C}{4 A^2} \end{align*}

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

\begin{align*} r &= \frac {-1}{1}\tag {6} \end{align*}

Comparing the above to (5) shows that

\begin{align*} s &= -1\\ t &= 1 \end{align*}

Therefore eq. (4) becomes

\begin{align*} z''(x) &= -z \left (x \right ) \tag {7} \end{align*}

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

\begin{align*} y &= z \left (x \right ) e^{-\int \frac {B}{2 A} \,dx} \end{align*}

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 \(\infty \). The following table summarizes these cases.

Case	Allowed pole order for \(r\)	Allowed value for \(\mathcal {O}(\infty )\)
1	\(\left \{ 0,1,2,4,6,8,\cdots \right \} \)	\(\left \{ \cdots ,-6,-4,-2,0,2,3,4,5,6,\cdots \right \} \)
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	\(\left \{ 1,2\right \} \)	\(\left \{ 2,3,4,5,6,7,\cdots \right \} \)

Table 2.23: Necessary conditions for each Kovacic case

The order of \(r\) at \(\infty \) is the degree of \(t\) minus the degree of \(s\). Therefore

\begin{align*} O\left (\infty \right ) &= \text {deg}(t) - \text {deg}(s) \\ &= 0 - 0 \\ &= 0 \end{align*}

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

\begin{align*} L &= [1] \end{align*}

Since \(r = -1\) is not a function of \(x\), then there is no need run Kovacic algorithm to obtain a solution for transformed ode \(z''=r z\) as one solution is

\[ z_1(x) = \cos \left (x \right ) \]

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

\[ y_1 = z_1 e^{ \int -\frac {1}{2} \frac {B}{A} \,dx} \]

Since \(B=0\) then the above reduces to

\begin{align*} y_1 &= z_1 \\ &= \cos \left (x \right ) \\ \end{align*}

Which simplifies to

\[ y_1 = \cos \left (x \right ) \]

The second solution \(y_2\) to the original ode is found using reduction of order

\[ y_2 = y_1 \int \frac { e^{\int -\frac {B}{A} \,dx}}{y_1^2} \,dx \]

Since \(B=0\) then the above becomes

\begin{align*} y_2 &= y_1 \int \frac {1}{y_1^2} \,dx \\ &= \cos \left (x \right )\int \frac {1}{\cos \left (x \right )^{2}} \,dx \\ &= \cos \left (x \right )\left (\tan \left (x \right )\right ) \\ \end{align*}

Therefore the solution is

\begin{align*} y &= c_1 y_1 + c_2 y_2 \\ &= c_1 \left (\cos \left (x \right )\right ) + c_2 \left (\cos \left (x \right )\left (\tan \left (x \right )\right )\right ) \\ \end{align*}

This is second order nonhomogeneous ODE. Let the solution be

\[ y = y_h + y_p \]

Where \(y_h\) is the solution to the homogeneous ODE \( A y''(x) + B y'(x) + C y(x) = 0\), and \(y_p\) is a particular solution to the nonhomogeneous ODE \(A y''(x) + B y'(x) + C y(x) = f(x)\). \(y_h\) is the solution to

\[ y^{\prime \prime }+y = 0 \]

The homogeneous solution is found using the Kovacic algorithm which results in

\[ y_h = c_1 \cos \left (x \right )+c_2 \sin \left (x \right ) \]

The particular solution is now found using the method of undetermined coefficients. Looking at the RHS of the ode, which is

\[ 1 \]

Shows that the corresponding undetermined set of the basis functions (UC_set) for the trial solution is

\[ [\{1\}] \]

While the set of the basis functions for the homogeneous solution found earlier is

\[ \{\cos \left (x \right ), \sin \left (x \right )\} \]

Since there is no duplication between the basis function in the UC_set and the basis functions of the homogeneous solution, the trial solution is a linear combination of all the basis in the UC_set.

\[ y_p = A_{1} \]

The unknowns \(\{A_{1}\}\) are found by substituting the above trial solution \(y_p\) into the ODE and comparing coefficients. Substituting the trial solution into the ODE and simplifying gives

\[ A_{1} = 1 \]

Solving for the unknowns by comparing coefficients results in

\[ [A_{1} = 1] \]

Substituting the above back in the above trial solution \(y_p\), gives the particular solution

\[ y_p = 1 \]

Therefore the general solution is

\begin{align*} y &= y_h + y_p \\ &= \left (c_1 \cos \left (x \right )+c_2 \sin \left (x \right )\right ) + \left (1\right ) \\ \end{align*}

Will add steps showing solving for IC soon.

Summary of solutions found

\begin{align*} y &= 1+c_1 \cos \left (x \right )+c_2 \sin \left (x \right ) \\ \end{align*}

Figure 2.107: Slope field \(y^{\prime \prime }+y = 1\)

✓ Maple. Time used: 0.002 (sec). Leaf size: 14

ode:=diff(diff(y(x),x),x)+y(x) = 1; 
dsolve(ode,y(x), singsol=all);
 

\[ y = \sin \left (x \right ) c_2 +\cos \left (x \right ) c_1 +1 \]

Maple trace

Methods for second order ODEs: 
--- Trying classification methods --- 
trying a quadrature 
trying high order exact linear fully integrable 
trying differential order: 2; linear nonhomogeneous with symmetry [0,1] 
trying a double symmetry of the form [xi=0, eta=F(x)] 
-> Try solving first the homogeneous part of the ODE 
   checking if the LODE has constant coefficients 
   <- constant coefficients successful 
<- solving first the homogeneous part of the ODE successful
 

Maple step by step

\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & \frac {d}{d x}\frac {d}{d x}y \left (x \right )+y \left (x \right )=1 \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 2 \\ {} & {} & \frac {d}{d x}\frac {d}{d x}y \left (x \right ) \\ \bullet & {} & \textrm {Characteristic polynomial of homogeneous ODE}\hspace {3pt} \\ {} & {} & r^{2}+1=0 \\ \bullet & {} & \textrm {Use quadratic formula to solve for}\hspace {3pt} r \\ {} & {} & r =\frac {0\pm \left (\sqrt {-4}\right )}{2} \\ \bullet & {} & \textrm {Roots of the characteristic polynomial}\hspace {3pt} \\ {} & {} & r =\left (\mathrm {-I}, \mathrm {I}\right ) \\ \bullet & {} & \textrm {1st solution of the homogeneous ODE}\hspace {3pt} \\ {} & {} & y_{1}\left (x \right )=\cos \left (x \right ) \\ \bullet & {} & \textrm {2nd solution of the homogeneous ODE}\hspace {3pt} \\ {} & {} & y_{2}\left (x \right )=\sin \left (x \right ) \\ \bullet & {} & \textrm {General solution of the ODE}\hspace {3pt} \\ {} & {} & y \left (x \right )=\mathit {C1} y_{1}\left (x \right )+\mathit {C2} y_{2}\left (x \right )+y_{p}\left (x \right ) \\ \bullet & {} & \textrm {Substitute in solutions of the homogeneous ODE}\hspace {3pt} \\ {} & {} & y \left (x \right )=\mathit {C1} \cos \left (x \right )+\mathit {C2} \sin \left (x \right )+y_{p}\left (x \right ) \\ \square & {} & \textrm {Find a particular solution}\hspace {3pt} y_{p}\left (x \right )\hspace {3pt}\textrm {of the ODE}\hspace {3pt} \\ {} & \circ & \textrm {Use variation of parameters to find}\hspace {3pt} y_{p}\hspace {3pt}\textrm {here}\hspace {3pt} f \left (x \right )\hspace {3pt}\textrm {is the forcing function}\hspace {3pt} \\ {} & {} & \left [y_{p}\left (x \right )=-y_{1}\left (x \right ) \int \frac {y_{2}\left (x \right ) f \left (x \right )}{W \left (y_{1}\left (x \right ), y_{2}\left (x \right )\right )}d x +y_{2}\left (x \right ) \int \frac {y_{1}\left (x \right ) f \left (x \right )}{W \left (y_{1}\left (x \right ), y_{2}\left (x \right )\right )}d x , f \left (x \right )=1\right ] \\ {} & \circ & \textrm {Wronskian of solutions of the homogeneous equation}\hspace {3pt} \\ {} & {} & W \left (y_{1}\left (x \right ), y_{2}\left (x \right )\right )=\left [\begin {array}{cc} \cos \left (x \right ) & \sin \left (x \right ) \\ -\sin \left (x \right ) & \cos \left (x \right ) \end {array}\right ] \\ {} & \circ & \textrm {Compute Wronskian}\hspace {3pt} \\ {} & {} & W \left (y_{1}\left (x \right ), y_{2}\left (x \right )\right )=1 \\ {} & \circ & \textrm {Substitute functions into equation for}\hspace {3pt} y_{p}\left (x \right ) \\ {} & {} & y_{p}\left (x \right )=-\cos \left (x \right ) \int \sin \left (x \right )d x +\sin \left (x \right ) \int \cos \left (x \right )d x \\ {} & \circ & \textrm {Compute integrals}\hspace {3pt} \\ {} & {} & y_{p}\left (x \right )=1 \\ \bullet & {} & \textrm {Substitute particular solution into general solution to ODE}\hspace {3pt} \\ {} & {} & y \left (x \right )=\mathit {C1} \cos \left (x \right )+\mathit {C2} \sin \left (x \right )+1 \end {array} \]

✓ Mathematica. Time used: 0.012 (sec). Leaf size: 17

ode=D[y[x],{x,2}]+y[x]==1; 
ic={}; 
DSolve[{ode,ic},y[x],x,IncludeSingularSolutions->True]
 

\[ y(x)\to c_1 \cos (x)+c_2 \sin (x)+1 \]

✓ Sympy. Time used: 0.059 (sec). Leaf size: 14

from sympy import * 
x = symbols("x") 
y = Function("y") 
ode = Eq(y(x) + Derivative(y(x), (x, 2)) - 1,0) 
ics = {} 
dsolve(ode,func=y(x),ics=ics)
 

\[ y{\left (x \right )} = C_{1} \sin {\left (x \right )} + C_{2} \cos {\left (x \right )} + 1 \]