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2.1.63 Problem 63

Solved using first_order_ode_homog_type_C
✓Maple
✓Mathematica
✗Sympy

Internal problem ID [10321]
Book : First order enumerated odes
Section : section 1
Problem number : 63
Date solved : Thursday, November 27, 2025 at 10:33:39 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 (a +x b +y\right )^{4} \\ \end{align*}
Let
\begin{align*} z = a +x b +y\tag {1} \end{align*}

Then

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

Therefore

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

Hence the given ode can now be written as

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

This is separable first order ode. Integrating

\begin{align*} \int d x&=\int \frac {1}{z^{4}+b}d z \\ x +c_1&=\frac {\sqrt {2}\, \left (\ln \left (\frac {z^{2}+b^{{1}/{4}} z \sqrt {2}+\sqrt {b}}{z^{2}-b^{{1}/{4}} z \sqrt {2}+\sqrt {b}}\right )+2 \arctan \left (\frac {\sqrt {2}\, z}{b^{{1}/{4}}}+1\right )+2 \arctan \left (\frac {\sqrt {2}\, z}{b^{{1}/{4}}}-1\right )\right )}{8 b^{{3}/{4}}} \\ \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 {\sqrt {2}\, b^{{5}/{4}} x +\sqrt {2}\, \left (a +y\right ) b^{{1}/{4}}+\sqrt {b}+y^{2}+\left (2 x b +2 a \right ) y+b^{2} x^{2}+2 a b x +a^{2}}{-\sqrt {2}\, b^{{5}/{4}} x -\sqrt {2}\, \left (a +y\right ) b^{{1}/{4}}+\sqrt {b}+y^{2}+\left (2 x b +2 a \right ) y+b^{2} x^{2}+2 a b x +a^{2}}\right )+2 \arctan \left (\frac {\sqrt {2}\, \left (a +x b +y\right )}{b^{{1}/{4}}}+1\right )+2 \arctan \left (\frac {\sqrt {2}\, \left (a +x b +y\right )}{b^{{1}/{4}}}-1\right )\right )}{8 b^{{3}/{4}}} &= x +c_1 \\ \end{align*}

Summary of solutions found

\begin{align*} \frac {\sqrt {2}\, \left (\ln \left (\frac {\sqrt {2}\, b^{{5}/{4}} x +\sqrt {2}\, \left (a +y\right ) b^{{1}/{4}}+\sqrt {b}+y^{2}+\left (2 x b +2 a \right ) y+b^{2} x^{2}+2 a b x +a^{2}}{-\sqrt {2}\, b^{{5}/{4}} x -\sqrt {2}\, \left (a +y\right ) b^{{1}/{4}}+\sqrt {b}+y^{2}+\left (2 x b +2 a \right ) y+b^{2} x^{2}+2 a b x +a^{2}}\right )+2 \arctan \left (\frac {\sqrt {2}\, \left (a +x b +y\right )}{b^{{1}/{4}}}+1\right )+2 \arctan \left (\frac {\sqrt {2}\, \left (a +x b +y\right )}{b^{{1}/{4}}}-1\right )\right )}{8 b^{{3}/{4}}} &= x +c_1 \\ \end{align*}
✓ Maple. Time used: 0.010 (sec). Leaf size: 49
ode:=diff(y(x),x) = (a+b*x+y(x))^4; 
dsolve(ode,y(x), singsol=all);
\[ y = -b x +\operatorname {RootOf}\left (-x +\int _{}^{\textit {\_Z}}\frac {1}{\textit {\_a}^{4}+4 \textit {\_a}^{3} a +6 \textit {\_a}^{2} a^{2}+4 \textit {\_a} \,a^{3}+a^{4}+b}d \textit {\_a} +c_1 \right ) \]

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 homogeneous C 
1st order, trying the canonical coordinates of the invariance group 
   -> Calling odsolve with the ODE, diff(y(x),x) = -b, 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

\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & \frac {d}{d x}y \left (x \right )=\left (a +b x +y \left (x \right )\right )^{4} \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 1 \\ {} & {} & \frac {d}{d x}y \left (x \right ) \\ \bullet & {} & \textrm {Solve for the highest derivative}\hspace {3pt} \\ {} & {} & \frac {d}{d x}y \left (x \right )=\left (a +b x +y \left (x \right )\right )^{4} \end {array} \]
✓ Mathematica. Time used: 0.3 (sec). Leaf size: 523
ode=D[y[x],x]==(a+b*x+y[x])^(4); 
ic={}; 
DSolve[{ode,ic},y[x],x,IncludeSingularSolutions->True]
\[ \text {Solve}\left [\int _1^{y(x)}\left (-\int _1^x\frac {b \left (4 a^3+12 b K[1] a^2+12 K[2] a^2+12 b^2 K[1]^2 a+12 K[2]^2 a+24 b K[1] K[2] a+4 b^3 K[1]^3+4 K[2]^3+12 b K[1] K[2]^2+12 b^2 K[1]^2 K[2]\right )}{\left (a^4+4 b K[1] a^3+4 K[2] a^3+6 b^2 K[1]^2 a^2+6 K[2]^2 a^2+12 b K[1] K[2] a^2+4 b^3 K[1]^3 a+4 K[2]^3 a+12 b K[1] K[2]^2 a+12 b^2 K[1]^2 K[2] a+b^4 K[1]^4+K[2]^4+4 b K[1] K[2]^3+6 b^2 K[1]^2 K[2]^2+b+4 b^3 K[1]^3 K[2]\right )^2}dK[1]-\frac {1}{a^4+4 b x a^3+4 K[2] a^3+6 b^2 x^2 a^2+6 K[2]^2 a^2+12 b x K[2] a^2+4 b^3 x^3 a+4 K[2]^3 a+12 b x K[2]^2 a+12 b^2 x^2 K[2] a+b^4 x^4+K[2]^4+4 b x K[2]^3+6 b^2 x^2 K[2]^2+b+4 b^3 x^3 K[2]}\right )dK[2]+\int _1^x\left (1-\frac {b}{a^4+4 b K[1] a^3+4 y(x) a^3+6 b^2 K[1]^2 a^2+6 y(x)^2 a^2+12 b K[1] y(x) a^2+4 b^3 K[1]^3 a+4 y(x)^3 a+12 b K[1] y(x)^2 a+12 b^2 K[1]^2 y(x) a+b^4 K[1]^4+y(x)^4+4 b K[1] y(x)^3+6 b^2 K[1]^2 y(x)^2+b+4 b^3 K[1]^3 y(x)}\right )dK[1]=c_1,y(x)\right ] \]
✗ Sympy
from sympy import * 
x = symbols("x") 
a = symbols("a") 
b = symbols("b") 
y = Function("y") 
ode = Eq(-(a + b*x + y(x))**4 + Derivative(y(x), x),0) 
ics = {} 
dsolve(ode,func=y(x),ics=ics)
 

TypeError : argument of type Mul is not iterable

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