2.1.69 Problem 69

Internal problem ID [9052]
Book : First order enumerated odes
Section : section 1
Problem number : 69
Date solved : Friday, April 25, 2025 at 05:39:30 PM
CAS classification : [[_1st_order, `_with_symmetry_[F(x),G(x)]`]]

Solved using first_order_ode_special_form_ID_1

Time used: 0.363 (sec)

Solve

$$\begin{array}{rl}{y}^{\prime }& =x{\mathrm{e}}^{x+y}+\sin \left(x\right)\end{array}$$

Writing the ode as

$$\begin{array}{}\text{(1)}& y^{\prime }& =x{\mathrm{e}}^{x+y}+\sin \left(x\right)\end{array}$$

And using the substitution $u={\mathrm{e}}^{-y}$ then

$$\begin{array}{rl}{u}^{\prime }& =-y^{\prime }{\mathrm{e}}^{-y}\end{array}$$

The above shows that

$$\begin{array}{rl}{y}^{\prime }& =-u^{\prime }\left(x\right){\mathrm{e}}^y\\ & =-\frac{u^{\prime }\left(x\right)}{u}\end{array}$$

Substituting this in (1) gives

$$\begin{array}{rl}-\frac{u^{\prime }\left(x\right)}{u}& =\frac{x{\mathrm{e}}^x}{u}+\sin \left(x\right)\end{array}$$

The above simplifies to

$$\begin{array}{rl}-u^{\prime }\left(x\right)& =x{\mathrm{e}}^x+\sin \left(x\right)u\left(x\right)\\ \text{(2)}& u^{\prime }\left(x\right)+\sin \left(x\right)u\left(x\right)& =-x{\mathrm{e}}^x\end{array}$$

Now ode (2) is solved for $u\left(x\right)$.

In canonical form a linear first order is

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

Comparing the above to the given ode shows that

$$\begin{array}{rl}q(x)& =\sin \left(x\right)\\ p(x)& =-x{\mathrm{e}}^x\end{array}$$

The integrating factor $\mu$ is

$$\begin{array}{rl}\mu & =e^{\int qdx}\\ & ={\mathrm{e}}^{\int \sin \left(x\right)dx}\\ & ={\mathrm{e}}^{-\cos \left(x\right)}\end{array}$$

The ode becomes

$$\begin{array}{rl}\frac{\mathrm{d}}{\mathrm{d}x}\left(\mu u\right)& =\mu p\\ \frac{\mathrm{d}}{\mathrm{d}x}\left(\mu u\right)& =\left(\mu \right)(-x{\mathrm{e}}^x)\\ \frac{\mathrm{d}}{\mathrm{d}x}\left(u{\mathrm{e}}^{-\cos \left(x\right)}\right)& =\left({\mathrm{e}}^{-\cos \left(x\right)}\right)(-x{\mathrm{e}}^x)\\ \mathrm{d}\left(u{\mathrm{e}}^{-\cos \left(x\right)}\right)& =(-x{\mathrm{e}}^x{\mathrm{e}}^{-\cos \left(x\right)})\mathrm{d}x\end{array}$$

Integrating gives

$$\begin{array}{rl}u{\mathrm{e}}^{-\cos \left(x\right)}& =\int -x{\mathrm{e}}^x{\mathrm{e}}^{-\cos \left(x\right)}dx\\ & =\int -x{\mathrm{e}}^x{\mathrm{e}}^{-\cos \left(x\right)}dx+c_1\end{array}$$

Dividing throughout by the integrating factor ${\mathrm{e}}^{-\cos \left(x\right)}$ gives the final solution

$$u\left(x\right)={\mathrm{e}}^{\cos \left(x\right)}(\int -x{\mathrm{e}}^x{\mathrm{e}}^{-\cos \left(x\right)}dx+c_1)$$

Substituting the solution found for $u\left(x\right)$ in $u={\mathrm{e}}^{-y}$ gives

$$\begin{array}{r}{\mathrm{e}}^{-y}={\mathrm{e}}^{\cos \left(x\right)}(\int -x{\mathrm{e}}^x{\mathrm{e}}^{-\cos \left(x\right)}dx+c_1)\end{array}$$

Solving for $y$ gives

$$\begin{array}{rl}y& =-\ln (-\int x{\mathrm{e}}^x{\mathrm{e}}^{-\cos \left(x\right)}dx+c_1)-\cos \left(x\right)\end{array}$$

Figure 2.74: Slope field $y^{\prime }=x{\mathrm{e}}^{x+y}+\sin \left(x\right)$

Summary of solutions found

$$\begin{array}{rl}y& =-\ln (-\int x{\mathrm{e}}^x{\mathrm{e}}^{-\cos \left(x\right)}dx+c_1)-\cos \left(x\right)\end{array}$$

✓ Maple. Time used: 0.010 (sec). Leaf size: 29

ode:=diff(y(x),x) = x*exp(x+y(x))+sin(x); 
dsolve(ode,y(x), singsol=all);
 

$$y=-\cos \left(x\right)-\ln (-c_1-\int x{\mathrm{e}}^{x-\cos \left(x\right)}dx)$$

Maple trace

Methods for first order ODEs: 
--- Trying classification methods --- 
trying a quadrature 
trying 1st order linear 
trying Bernoulli 
trying separable 
trying inverse linear 
trying homogeneous types: 
trying Chini 
differential order: 1; looking for linear symmetries 
trying exact 
Looking for potential symmetries 
trying inverse_Riccati 
trying an equivalence to an Abel ODE 
differential order: 1; trying a linearization to 2nd order 
--- trying a change of variables {x -> y(x), y(x) -> x} 
differential order: 1; trying a linearization to 2nd order 
trying 1st order ODE linearizable_by_differentiation 
--- Trying Lie symmetry methods, 1st order --- 
   -> Computing symmetries using: way = 3 
   -> Computing symmetries using: way = 4 
   -> Computing symmetries using: way = 5 
trying symmetry patterns for 1st order ODEs 
-> trying a symmetry pattern of the form [F(x)*G(y), 0] 
-> trying a symmetry pattern of the form [0, F(x)*G(y)] 
<- symmetry pattern of the form [0, F(x)*G(y)] successful
 

Maple step by step

$$\begin{array}{lll}& & \text{Let’s solve}\\ & & \frac{d}{dx}y\left(x\right)=x{\mathrm{e}}^{x+y\left(x\right)}+\sin \left(x\right)\\ \bullet & & \text{Highest derivative means the order of the ODE is}1\\ & & \frac{d}{dx}y\left(x\right)\\ \bullet & & \text{Solve for the highest derivative}\\ & & \frac{d}{dx}y\left(x\right)=x{\mathrm{e}}^{x+y\left(x\right)}+\sin \left(x\right)\end{array}$$

✓ Mathematica. Time used: 3.158 (sec). Leaf size: 100

ode=D[y[x],x]==x*Exp[x+y[x]]+Sin[x]; 
ic={}; 
DSolve[{ode,ic},y[x],x,IncludeSingularSolutions->True]
 

$$\text{Solve}[{\int }_1^x(-e^{K[1]-\cos (K[1])}K[1]-e^{-\cos (K[1])-y(x)}\sin (K[1]))dK[1]+{\int }_1^{y(x)}-e^{-\cos (x)-K[2]}(e^{\cos (x)+K[2]}{\int }_1^x{e}^{-\cos (K[1])-K[2]}\sin (K[1])dK[1]-1)dK[2]=c_1,y(x)]$$

✓ Sympy. Time used: 16.325 (sec). Leaf size: 24

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

$$y\left(x\right)=\log \left(\frac{e^{-\cos \left(x\right)}}{C_1-\int x{e}^x{e}^{-\cos \left(x\right)}dx}\right)$$