2.2.7 Problem 8

Internal problem ID [10418]
Book : Second order enumerated odes
Section : section 2
Problem number : 8
Date solved : Thursday, November 27, 2025 at 10:51:25 AM
CAS classification : [[_homogeneous, `class C`], _dAlembert]

Solved using first_order_ode_homog_type_C

Time used: 0.035 (sec)

Solve

\begin{align*} y^{\prime }&=\left (x +y\right )^{4} \\ \end{align*}

Let

\begin{align*} z = x +y\tag {1} \end{align*}

Then

\begin{align*} z^{\prime }\left (x \right )&=y^{\prime }+1 \end{align*}

Therefore

\begin{align*} y^{\prime }&=z^{\prime }\left (x \right )-1 \end{align*}

Hence the given ode can now be written as

\begin{align*} z^{\prime }\left (x \right )-1&=z^{4} \end{align*}

This is separable first order ode. Integrating

\begin{align*} \int d x&=\int \frac {1}{z^{4}+1}d z \\ x +c_1&=\frac {\sqrt {2}\, \left (\ln \left (\frac {z^{2}+\sqrt {2}\, z +1}{z^{2}-\sqrt {2}\, z +1}\right )+2 \arctan \left (\sqrt {2}\, z +1\right )+2 \arctan \left (\sqrt {2}\, z -1\right )\right )}{8} \\ \end{align*}

Replacing \(z\) back by its value from (1) then the above gives the solution as Simplifying the above gives

\begin{align*} \frac {\sqrt {2}\, \left (\ln \left (\frac {\left (x +y\right ) \sqrt {2}+x^{2}+2 x y+y^{2}+1}{\left (-y-x \right ) \sqrt {2}+x^{2}+2 x y+y^{2}+1}\right )+2 \arctan \left (\left (x +y\right ) \sqrt {2}+1\right )+2 \arctan \left (\left (x +y\right ) \sqrt {2}-1\right )\right )}{8} &= x +c_1 \\ \end{align*}

Figure 2.120: Slope field \(y^{\prime } = \left (x +y\right )^{4}\)

Summary of solutions found

\begin{align*} \frac {\sqrt {2}\, \left (\ln \left (\frac {\left (x +y\right ) \sqrt {2}+x^{2}+2 x y+y^{2}+1}{\left (-y-x \right ) \sqrt {2}+x^{2}+2 x y+y^{2}+1}\right )+2 \arctan \left (\left (x +y\right ) \sqrt {2}+1\right )+2 \arctan \left (\left (x +y\right ) \sqrt {2}-1\right )\right )}{8} &= x +c_1 \\ \end{align*}

✓ Maple. Time used: 0.264 (sec). Leaf size: 882

ode:=diff(y(x),x) = (x+y(x))^4; 
dsolve(ode,y(x), singsol=all);
 

\[ \text {Expression too large to display} \]

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 homogeneous C 
1st order, trying the canonical coordinates of the invariance group 
   -> Calling odsolve with the ODE, diff(y(x),x) = -1, y(x) 
      *** Sublevel 2 *** 
      Methods for first order ODEs: 
      --- Trying classification methods --- 
      trying a quadrature 
      trying 1st order linear 
      <- 1st order linear successful 
<- 1st order, canonical coordinates successful 
<- homogeneous successful
 

Maple step by step

\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & \frac {d}{d x}y \left (x \right )=\left (x +y \left (x \right )\right )^{4} \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 1 \\ {} & {} & \frac {d}{d x}y \left (x \right ) \\ \bullet & {} & \textrm {Solve for the highest derivative}\hspace {3pt} \\ {} & {} & \frac {d}{d x}y \left (x \right )=\left (x +y \left (x \right )\right )^{4} \end {array} \]

✓ Mathematica. Time used: 0.087 (sec). Leaf size: 175

ode=D[y[x],x] == (x + y[x])^4; 
ic={}; 
DSolve[{ode,ic},y[x],x,IncludeSingularSolutions->True]
 

\[ \text {Solve}\left [\int _1^{y(x)}\left (\frac {1}{x^4+4 K[2] x^3+6 K[2]^2 x^2+4 K[2]^3 x+K[2]^4+1}-\int _1^x-\frac {4 K[1]^3+12 K[2] K[1]^2+12 K[2]^2 K[1]+4 K[2]^3}{\left (K[1]^4+4 K[2] K[1]^3+6 K[2]^2 K[1]^2+4 K[2]^3 K[1]+K[2]^4+1\right )^2}dK[1]\right )dK[2]+\int _1^x\left (\frac {1}{K[1]^4+4 y(x) K[1]^3+6 y(x)^2 K[1]^2+4 y(x)^3 K[1]+y(x)^4+1}-1\right )dK[1]=c_1,y(x)\right ] \]

✓ Sympy. Time used: 1.692 (sec). Leaf size: 109

from sympy import * 
x = symbols("x") 
y = Function("y") 
ode = Eq(-(x + y(x))**4 + Derivative(y(x), x),0) 
ics = {} 
dsolve(ode,func=y(x),ics=ics)
 

\[ C_{1} + x - \frac {\sqrt {2} \log {\left (\left (- x - y{\left (x \right )}\right )^{2} - \sqrt {2} \left (- x - y{\left (x \right )}\right ) + 1 \right )}}{8} + \frac {\sqrt {2} \log {\left (\left (- x - y{\left (x \right )}\right )^{2} + \sqrt {2} \left (- x - y{\left (x \right )}\right ) + 1 \right )}}{8} + \frac {\sqrt {2} \operatorname {atan}{\left (\sqrt {2} \left (- x - y{\left (x \right )}\right ) - 1 \right )}}{4} + \frac {\sqrt {2} \operatorname {atan}{\left (\sqrt {2} \left (- x - y{\left (x \right )}\right ) + 1 \right )}}{4} = 0 \]