2.1.68 Problem 68

Internal problem ID [10326]
Book : First order enumerated odes
Section : section 1
Problem number : 68
Date solved : Thursday, November 27, 2025 at 10:34:11 AM
CAS classification : [[_1st_order, `_with_symmetry_[F(x),G(x)]`]]

Solved using first_order_ode_special_form_ID_1

Time used: 0.135 (sec)

Solve

\begin{align*} y^{\prime }&=10 \,{\mathrm e}^{x +y}+x^{2} \\ \end{align*}

Writing the ode as

\begin{align*} y^{\prime } &= 10 \,{\mathrm e}^{x +y}+x^{2}\tag {1} \end{align*}

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

\begin{align*} u' &= -y^{\prime } {\mathrm e}^{-y} \end{align*}

The above shows that

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

Substituting this in (1) gives

\begin{align*} -\frac {u^{\prime }\left (x \right )}{u}&=\frac {10 \,{\mathrm e}^{x}}{u}+x^{2} \end{align*}

The above simplifies to

\begin{align*} -u^{\prime }\left (x \right )&=10 \,{\mathrm e}^{x}+x^{2} u \left (x \right )\\ u^{\prime }\left (x \right )+x^{2} u \left (x \right )&=-10 \,{\mathrm e}^{x}\tag {2} \end{align*}

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

In canonical form a linear first order is

\begin{align*} u^{\prime }\left (x \right ) + q(x)u \left (x \right ) &= p(x) \end{align*}

Comparing the above to the given ode shows that

\begin{align*} q(x) &=x^{2}\\ p(x) &=-10 \,{\mathrm e}^{x} \end{align*}

The integrating factor \(\mu \) is

\begin{align*} \mu &= e^{\int {q\,dx}}\\ &= {\mathrm e}^{\int x^{2}d x}\\ &= {\mathrm e}^{\frac {x^{3}}{3}} \end{align*}

The ode becomes

\begin{align*} \frac {\mathop {\mathrm {d}}}{ \mathop {\mathrm {d}x}}\left ( \mu u\right ) &= \mu p \\ \frac {\mathop {\mathrm {d}}}{ \mathop {\mathrm {d}x}}\left ( \mu u\right ) &= \left (\mu \right ) \left (-10 \,{\mathrm e}^{x}\right ) \\ \frac {\mathop {\mathrm {d}}}{ \mathop {\mathrm {d}x}} \left (u \,{\mathrm e}^{\frac {x^{3}}{3}}\right ) &= \left ({\mathrm e}^{\frac {x^{3}}{3}}\right ) \left (-10 \,{\mathrm e}^{x}\right ) \\ \mathrm {d} \left (u \,{\mathrm e}^{\frac {x^{3}}{3}}\right ) &= \left (-10 \,{\mathrm e}^{x} {\mathrm e}^{\frac {x^{3}}{3}}\right )\, \mathrm {d} x \\ \end{align*}

Integrating gives

\begin{align*} u \,{\mathrm e}^{\frac {x^{3}}{3}}&= \int {-10 \,{\mathrm e}^{x} {\mathrm e}^{\frac {x^{3}}{3}} \,dx} \\ &=\int -10 \,{\mathrm e}^{x} {\mathrm e}^{\frac {x^{3}}{3}}d x + c_1 \end{align*}

Dividing throughout by the integrating factor \({\mathrm e}^{\frac {x^{3}}{3}}\) gives the final solution

\[ u \left (x \right ) = {\mathrm e}^{-\frac {x^{3}}{3}} \left (\int -10 \,{\mathrm e}^{x} {\mathrm e}^{\frac {x^{3}}{3}}d x +c_1 \right ) \]

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

\begin{align*} {\mathrm e}^{-y} = {\mathrm e}^{-\frac {x^{3}}{3}} \left (\int -10 \,{\mathrm e}^{x} {\mathrm e}^{\frac {x^{3}}{3}}d x +c_1 \right ) \end{align*}

Solving for \(y\) gives

\begin{align*} y &= -\ln \left (-10 \int {\mathrm e}^{x} {\mathrm e}^{\frac {x^{3}}{3}}d x +c_1 \right )+\frac {x^{3}}{3} \\ \end{align*}

Figure 2.78: Slope field \(y^{\prime } = 10 \,{\mathrm e}^{x +y}+x^{2}\)

Summary of solutions found

\begin{align*} y &= -\ln \left (-10 \int {\mathrm e}^{x} {\mathrm e}^{\frac {x^{3}}{3}}d x +c_1 \right )+\frac {x^{3}}{3} \\ \end{align*}

✓ Maple. Time used: 0.008 (sec). Leaf size: 30

ode:=diff(y(x),x) = 10*exp(x+y(x))+x^2; 
dsolve(ode,y(x), singsol=all);
 

\[ y = \frac {x^{3}}{3}-\ln \left (-c_1 -10 \int {\mathrm e}^{\frac {x \left (x^{2}+3\right )}{3}}d x \right ) \]

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

\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & \frac {d}{d x}y \left (x \right )=10 \,{\mathrm e}^{x +y \left (x \right )}+x^{2} \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 1 \\ {} & {} & \frac {d}{d x}y \left (x \right ) \\ \bullet & {} & \textrm {Solve for the highest derivative}\hspace {3pt} \\ {} & {} & \frac {d}{d x}y \left (x \right )=10 \,{\mathrm e}^{x +y \left (x \right )}+x^{2} \end {array} \]

✓ Mathematica. Time used: 0.253 (sec). Leaf size: 115

ode=D[y[x],x]==10*Exp[x+y[x]]+x^2; 
ic={}; 
DSolve[{ode,ic},y[x],x,IncludeSingularSolutions->True]
 

\[ \text {Solve}\left [\int _1^{y(x)}-\frac {1}{10} e^{-K[2]} \left (10 e^{K[2]} \int _1^x-\frac {1}{10} e^{\frac {K[1]^3}{3}-K[2]} K[1]^2dK[1]+e^{\frac {x^3}{3}}\right )dK[2]+\int _1^x\left (\frac {1}{10} e^{\frac {K[1]^3}{3}-y(x)} K[1]^2+e^{\frac {K[1]^3}{3}+K[1]}\right )dK[1]=c_1,y(x)\right ] \]

✓ Sympy. Time used: 0.963 (sec). Leaf size: 29

from sympy import * 
x = symbols("x") 
y = Function("y") 
ode = Eq(-x**2 - 10*exp(x + y(x)) + Derivative(y(x), x),0) 
ics = {} 
dsolve(ode,func=y(x),ics=ics)
 

\[ y{\left (x \right )} = \log {\left (\frac {\sqrt [3]{e^{x^{3}}}}{C_{1} - 10 \int e^{x} \sqrt [3]{e^{x^{3}}}\, dx} \right )} \]