2.1.31 Problem 31

Internal problem ID [9102]
Book : Second order enumerated odes
Section : section 1
Problem number : 31
Date solved : Sunday, March 30, 2025 at 02:07:06 PM
CAS classification : [[_2nd_order, _missing_x]]

Solved as second order linear constant coeff ode

Time used: 0.108 (sec)

Solve

$$\begin{array}{rl}{y}^{\prime \prime }+y^{\prime }& =1\end{array}$$

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

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

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

$$y=y_h+y_p$$

Where $y_h$ is the solution to the homogeneous ODE $A{y}^{″}(x)+B{y}^{\prime }(x)+Cy(x)=0$, and $y_p$ is a particular solution to the non-homogeneous ODE $A{y}^{″}(x)+B{y}^{\prime }(x)+Cy(x)=f(x)$. $y_h$ is the solution to

$$y^{\prime \prime }+y^{\prime }=0$$

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

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

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

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

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 $A=1,B=1,C=0$ into the above gives

$$\begin{array}{rl}{\lambda }_{1,2}& =\frac{-1}{(2)\left(1\right)}\pm \frac{1}{(2)\left(1\right)}\sqrt{1^2-(4)\left(1\right)\left(0\right)}\\ & =-\frac{1}{2}\pm \frac{1}{2}\end{array}$$

Hence

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

Which simplifies to

$$\begin{array}{rl}{\lambda }_1& =0\\ {\lambda }_2& =-1\end{array}$$

Since roots are real and distinct, then the solution is

$$\begin{array}{rl}y& =c_1{e}^{{\lambda }_{1}x}+c_2{e}^{{\lambda }_{2}x}\\ y& =c_1{e}^{\left(0\right)x}+c_2{e}^{(-1)x}\end{array}$$

Or

$$y=c_1+c_2{\mathrm{e}}^{-x}$$

Therefore the homogeneous solution $y_h$ is

$$y_h=c_1+c_2{\mathrm{e}}^{-x}$$

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

$$\{1,{\mathrm{e}}^{-x}\}$$

Since $1$ is duplicated in the UC_set, then this basis is multiplied by extra $x$. The UC_set becomes

$$[\{x\}]$$

Since there was duplication between the basis functions in the UC_set and the basis functions of the homogeneous solution, the trial solution is a linear combination of all the basis function in the above updated UC_set.

$$y_p=A_{1}x$$

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

$$y_p=x$$

Therefore the general solution is

$$\begin{array}{rl}y& =y_h+y_p\\ & =(c_1+c_2{\mathrm{e}}^{-x})+\left(x\right)\end{array}$$

Will add steps showing solving for IC soon.

Summary of solutions found

$$\begin{array}{rl}y& =x+c_1+c_2{\mathrm{e}}^{-x}\end{array}$$

Figure 2.63: Slope field $y^{\prime \prime }+y^{\prime }=1$

Solved as second order linear exact ode

Time used: 0.124 (sec)

Solve

$$\begin{array}{rl}{y}^{\prime \prime }+y^{\prime }& =1\end{array}$$

An ode of the form

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

is exact if

$$\begin{array}{}\text{(1)}& p^{″}(x)-q^{\prime }(x)+r(x)& =0\end{array}$$

For the given ode we have

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

Hence

$$\begin{array}{rl}{p}^{″}(x)& =0\\ q^{\prime }(x)& =0\end{array}$$

Therefore (1) becomes

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

Hence the ode is exact. Since we now know the ode is exact, it can be written as

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

Integrating gives

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

Substituting the above values for $p,q,r,s$ gives

$$\begin{array}{rl}{y}^{\prime }+y& =\int 1dx\end{array}$$

We now have a first order ode to solve which is

$$\begin{array}{r}{y}^{\prime }+y=x+c_1\end{array}$$

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

$$\begin{array}{rl}q(x)& =1\\ p(x)& =x+c_1\end{array}$$

The integrating factor $\mu$ is

$$\begin{array}{rl}\mu & =e^{\int qdx}\\ & ={\mathrm{e}}^{\int 1dx}\\ & ={\mathrm{e}}^x\end{array}$$

The ode becomes

$$\begin{array}{rl}\frac{\mathrm{d}}{\mathrm{d}x}\left(\mu y\right)& =\mu p\\ \frac{\mathrm{d}}{\mathrm{d}x}\left(\mu y\right)& =\left(\mu \right)(x+c_1)\\ \frac{\mathrm{d}}{\mathrm{d}x}\left(y{\mathrm{e}}^x\right)& =\left({\mathrm{e}}^x\right)(x+c_1)\\ \mathrm{d}\left(y{\mathrm{e}}^x\right)& =\left((x+c_1){\mathrm{e}}^x\right)\mathrm{d}x\end{array}$$

Integrating gives

$$\begin{array}{rl}y{\mathrm{e}}^x& =\int (x+c_1){\mathrm{e}}^{x}dx\\ & =(x+c_1-1){\mathrm{e}}^x+c_2\end{array}$$

Dividing throughout by the integrating factor ${\mathrm{e}}^x$ gives the final solution

$$y=c_2{\mathrm{e}}^{-x}+x+c_1-1$$

Will add steps showing solving for IC soon.

Summary of solutions found

$$\begin{array}{rl}y& =c_2{\mathrm{e}}^{-x}+x+c_1-1\end{array}$$

Figure 2.64: Slope field $y^{\prime \prime }+y^{\prime }=1$

Solved as second order missing y ode

Time used: 0.284 (sec)

Solve

$$\begin{array}{rl}{y}^{\prime \prime }+y^{\prime }& =1\end{array}$$

This is second order ode with missing dependent variable $y$. Let

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

Then

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

Hence the ode becomes

$$\begin{array}{r}{u}^{\prime }\left(x\right)+u\left(x\right)-1=0\end{array}$$

Which is now solved for $u(x)$ as first order ode.

Integrating gives

$$\begin{array}{rl}\int \frac{1}{1-u}du& =dx\\ -\ln (-1+u)& =x+c_1\end{array}$$

Singular solutions are found by solving

$$\begin{array}{rl}1-u& =0\end{array}$$

for $u$. 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.

$$\begin{array}{r}u=1\end{array}$$

In summary, these are the solution found for $u(x)$

$$\begin{array}{rl}-\ln (-1+u)& =x+c_1\\ u& =1\end{array}$$

For solution $-\ln (-1+u)=x+c_1$, since $u=y^{\prime }\left(x\right)$ then we now have a new first order ode to solve which is

$$\begin{array}{r}-\ln (-1+y^{\prime }\left(x\right))=x+c_1\end{array}$$

Since the ode has the form $y^{\prime }=f(x)$, then we only need to integrate $f(x)$.

$$\begin{array}{rl}\int dy& =\int {\mathrm{e}}^{-x-c_1}+1dx\\ y& =x-{\mathrm{e}}^{-x-c_1}+c_2\end{array}$$

For solution $u\left(x\right)=1$, since $u=y^{\prime }$ then we now have a new first order ode to solve which is

$$\begin{array}{r}{y}^{\prime }=1\end{array}$$

Since the ode has the form $y^{\prime }=f(x)$, then we only need to integrate $f(x)$.

$$\begin{array}{rl}\int dy& =\int 1dx\\ y& =x+c_3\end{array}$$

In summary, these are the solution found for $(y)$

$$\begin{array}{rl}y& =x-{\mathrm{e}}^{-x-c_1}+c_2\\ y& =x+c_3\end{array}$$

Will add steps showing solving for IC soon.

Summary of solutions found

$$\begin{array}{rl}y& =x+c_3\\ y& =x-{\mathrm{e}}^{-x-c_1}+c_2\end{array}$$

Figure 2.65: Slope field $y^{\prime \prime }+y^{\prime }=1$

Solved as second order integrable as is ode

Time used: 0.070 (sec)

Solve

$$\begin{array}{rl}{y}^{\prime \prime }+y^{\prime }& =1\end{array}$$

Integrating both sides of the ODE w.r.t $x$ gives

$$\begin{array}{rl}\int (y^{\prime \prime }+y^{\prime })dx& =\int 1dx\\ y^{\prime }+y=x+c_1\end{array}$$

Which 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

$$\begin{array}{rl}q(x)& =1\\ p(x)& =x+c_1\end{array}$$

The integrating factor $\mu$ is

$$\begin{array}{rl}\mu & =e^{\int qdx}\\ & ={\mathrm{e}}^{\int 1dx}\\ & ={\mathrm{e}}^x\end{array}$$

The ode becomes

$$\begin{array}{rl}\frac{\mathrm{d}}{\mathrm{d}x}\left(\mu y\right)& =\mu p\\ \frac{\mathrm{d}}{\mathrm{d}x}\left(\mu y\right)& =\left(\mu \right)(x+c_1)\\ \frac{\mathrm{d}}{\mathrm{d}x}\left(y{\mathrm{e}}^x\right)& =\left({\mathrm{e}}^x\right)(x+c_1)\\ \mathrm{d}\left(y{\mathrm{e}}^x\right)& =\left((x+c_1){\mathrm{e}}^x\right)\mathrm{d}x\end{array}$$

Integrating gives

$$\begin{array}{rl}y{\mathrm{e}}^x& =\int (x+c_1){\mathrm{e}}^{x}dx\\ & =(x+c_1-1){\mathrm{e}}^x+c_2\end{array}$$

Dividing throughout by the integrating factor ${\mathrm{e}}^x$ gives the final solution

$$y=c_2{\mathrm{e}}^{-x}+x+c_1-1$$

Will add steps showing solving for IC soon.

Summary of solutions found

$$\begin{array}{rl}y& =c_2{\mathrm{e}}^{-x}+x+c_1-1\end{array}$$

Figure 2.66: Slope field $y^{\prime \prime }+y^{\prime }=1$

Solved as second order integrable as is ode (ABC method)

Time used: 0.060 (sec)

Solve

$$\begin{array}{rl}{y}^{\prime \prime }+y^{\prime }& =1\end{array}$$

Writing the ode as

$$y^{\prime \prime }+y^{\prime }=1$$

Integrating both sides of the ODE w.r.t $x$ gives

$$\begin{array}{rl}\int (y^{\prime \prime }+y^{\prime })dx& =\int 1dx\\ y^{\prime }+y=x+c_1\end{array}$$

Which 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

$$\begin{array}{rl}q(x)& =1\\ p(x)& =x+c_1\end{array}$$

The integrating factor $\mu$ is

$$\begin{array}{rl}\mu & =e^{\int qdx}\\ & ={\mathrm{e}}^{\int 1dx}\\ & ={\mathrm{e}}^x\end{array}$$

The ode becomes

$$\begin{array}{rl}\frac{\mathrm{d}}{\mathrm{d}x}\left(\mu y\right)& =\mu p\\ \frac{\mathrm{d}}{\mathrm{d}x}\left(\mu y\right)& =\left(\mu \right)(x+c_1)\\ \frac{\mathrm{d}}{\mathrm{d}x}\left(y{\mathrm{e}}^x\right)& =\left({\mathrm{e}}^x\right)(x+c_1)\\ \mathrm{d}\left(y{\mathrm{e}}^x\right)& =\left((x+c_1){\mathrm{e}}^x\right)\mathrm{d}x\end{array}$$

Integrating gives

$$\begin{array}{rl}y{\mathrm{e}}^x& =\int (x+c_1){\mathrm{e}}^{x}dx\\ & =(x+c_1-1){\mathrm{e}}^x+c_2\end{array}$$

Dividing throughout by the integrating factor ${\mathrm{e}}^x$ gives the final solution

$$y=c_2{\mathrm{e}}^{-x}+x+c_1-1$$

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

$$\begin{array}{rl}q(x)& =1\\ p(x)& =x+c_1\end{array}$$

The integrating factor $\mu$ is

$$\begin{array}{rl}\mu & =e^{\int qdx}\\ & ={\mathrm{e}}^{\int 1dx}\\ & ={\mathrm{e}}^x\end{array}$$

The ode becomes

$$\begin{array}{rl}\frac{\mathrm{d}}{\mathrm{d}x}\left(\mu y\right)& =\mu p\\ \frac{\mathrm{d}}{\mathrm{d}x}\left(\mu y\right)& =\left(\mu \right)(x+c_1)\\ \frac{\mathrm{d}}{\mathrm{d}x}\left(y{\mathrm{e}}^x\right)& =\left({\mathrm{e}}^x\right)(x+c_1)\\ \mathrm{d}\left(y{\mathrm{e}}^x\right)& =\left((x+c_1){\mathrm{e}}^x\right)\mathrm{d}x\end{array}$$

Integrating gives

$$\begin{array}{rl}y{\mathrm{e}}^x& =\int (x+c_1){\mathrm{e}}^{x}dx\\ & =(x+c_1-1){\mathrm{e}}^x+c_2\end{array}$$

Dividing throughout by the integrating factor ${\mathrm{e}}^x$ gives the final solution

$$y=c_2{\mathrm{e}}^{-x}+x+c_1-1$$

Will add steps showing solving for IC soon.

Figure 2.67: Slope field $y^{\prime \prime }+y^{\prime }=1$

Solved as second order ode using Kovacic algorithm

Time used: 0.161 (sec)

Solve

$$\begin{array}{rl}{y}^{\prime \prime }+y^{\prime }& =1\end{array}$$

Writing the ode as

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

Comparing (1) and (2) shows that

$$\begin{array}{rl}A& =1\\ \text{(3)}& B& =1C& =0\end{array}$$

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

$$\begin{array}{}\text{(6)}& r& =\frac{1}{4}\end{array}$$

Comparing the above to (5) shows that

$$\begin{array}{rl}s& =1\\ t& =4\end{array}$$

Therefore eq. (4) becomes

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

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 2.16: 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 $r=\frac{1}{4}$ 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

$$z_1(x)={\mathrm{e}}^{-\frac{x}{2}}$$

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

$$y_1={\mathrm{e}}^{-x}$$

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

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

Therefore the solution is

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

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 $A{y}^{″}(x)+B{y}^{\prime }(x)+Cy(x)=0$, and $y_p$ is a particular solution to the nonhomogeneous ODE $A{y}^{″}(x)+B{y}^{\prime }(x)+Cy(x)=f(x)$. $y_h$ is the solution to

$$y^{\prime \prime }+y^{\prime }=0$$

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

$$y_h=c_1{\mathrm{e}}^{-x}+c_2$$

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

$$\{1,{\mathrm{e}}^{-x}\}$$

Since $1$ is duplicated in the UC_set, then this basis is multiplied by extra $x$. The UC_set becomes

$$[\{x\}]$$

Since there was duplication between the basis functions in the UC_set and the basis functions of the homogeneous solution, the trial solution is a linear combination of all the basis function in the above updated UC_set.

$$y_p=A_{1}x$$

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

$$y_p=x$$

Therefore the general solution is

$$\begin{array}{rl}y& =y_h+y_p\\ & =(c_1{\mathrm{e}}^{-x}+c_2)+\left(x\right)\end{array}$$

Will add steps showing solving for IC soon.

Summary of solutions found

$$\begin{array}{rl}y& =x+c_1{\mathrm{e}}^{-x}+c_2\end{array}$$

Figure 2.68: Slope field $y^{\prime \prime }+y^{\prime }=1$

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

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

$$y=-{\mathrm{e}}^{-x}{c}_1+x+c_2$$

Maple trace

Methods for second order ODEs: 
--- Trying classification methods --- 
trying a quadrature 
trying high order exact linear fully integrable 
-> Calling odsolve with the ODE, diff(_b(_a),_a) = -_b(_a)+1, _b(_a) 
   *** Sublevel 2 *** 
   Methods for first order ODEs: 
   --- Trying classification methods --- 
   trying a quadrature 
   trying 1st order linear 
   <- 1st order linear successful 
<- high order exact linear fully integrable successful
 

Maple step by step

$$\begin{array}{lll}& & \text{Let’s solve}\\ & & \frac{d}{dx}\frac{d}{dx}y\left(x\right)+\frac{d}{dx}y\left(x\right)=1\\ \bullet & & \text{Highest derivative means the order of the ODE is}2\\ & & \frac{d}{dx}\frac{d}{dx}y\left(x\right)\\ \bullet & & \text{Characteristic polynomial of homogeneous ODE}\\ & & r^2+r=0\\ \bullet & & \text{Factor the characteristic polynomial}\\ & & r(r+1)=0\\ \bullet & & \text{Roots of the characteristic polynomial}\\ & & r=(-1,0)\\ \bullet & & \text{1st solution of the homogeneous ODE}\\ & & y_1\left(x\right)={\mathrm{e}}^{-x}\\ \bullet & & \text{2nd solution of the homogeneous ODE}\\ & & y_2\left(x\right)=1\\ \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)+y_p\left(x\right)\\ \bullet & & \text{Substitute in solutions of the homogeneous ODE}\\ & & y\left(x\right)=\mathit{C}\mathit{1}{\mathrm{e}}^{-x}+\mathit{C}\mathit{2}+y_p\left(x\right)\\ ◻& & \text{Find a particular solution}{y}_p\left(x\right)\text{of the ODE}\\ & \circ & \text{Use variation of parameters to find}{y}_p\text{here}f\left(x\right)\text{is the forcing function}\\ & & [y_p\left(x\right)=-y_1\left(x\right)\int \frac{y_2\left(x\right)f\left(x\right)}{W(y_1\left(x\right),y_2\left(x\right))}dx+y_2\left(x\right)\int \frac{y_1\left(x\right)f\left(x\right)}{W(y_1\left(x\right),y_2\left(x\right))}dx,f\left(x\right)=1]\\ & \circ & \text{Wronskian of solutions of the homogeneous equation}\\ & & W(y_1\left(x\right),y_2\left(x\right))=\left[\begin{array}{cc}{\mathrm{e}}^{-x}& 1\\ -{\mathrm{e}}^{-x}& 0\end{array}\right]\\ & \circ & \text{Compute Wronskian}\\ & & W(y_1\left(x\right),y_2\left(x\right))={\mathrm{e}}^{-x}\\ & \circ & \text{Substitute functions into equation for}{y}_p\left(x\right)\\ & & y_p\left(x\right)=-{\mathrm{e}}^{-x}\int {\mathrm{e}}^{x}dx+\int 1dx\\ & \circ & \text{Compute integrals}\\ & & y_p\left(x\right)=-1+x\\ \bullet & & \text{Substitute particular solution into general solution to ODE}\\ & & y\left(x\right)=\mathit{C}\mathit{1}{\mathrm{e}}^{-x}+\mathit{C}\mathit{2}-1+x\end{array}$$

✓ Mathematica. Time used: 0.013 (sec). Leaf size: 18

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

$$y(x)\to x-c_1{e}^{-x}+c_2$$

✓ Sympy. Time used: 0.118 (sec). Leaf size: 10

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

$$y\left(x\right)=C_1+C_2{e}^{-x}+x$$