2.4.7 Problem 7

Internal problem ID [10352]
Book : First order enumerated odes
Section : section 4. First order odes solved using series method
Problem number : 7
Date solved : Monday, December 08, 2025 at 08:06:21 PM
CAS classification : [_linear]

\begin{align*} y^{\prime } x +y&=2 x^{4}+x^{3}+x \\ \end{align*}

Series expansion around \(x=0\).

Entering first order ode series solverEntering first order ode series solver frobenius solverSince this is an inhomogeneous, then let the solution be

\[ y = y_h + y_p \]

Where \(y_h\) is the solution to the homogeneous ode \(y^{\prime } x +y = 0\),and \(y_p\) is a particular solution to the inhomogeneous ode. First, we solve for \(y_h\) Let the homogeneous solution be represented as Frobenius power series of the form

\[ y = \moverset {\infty }{\munderset {n =0}{\sum }}a_{n} x^{n +r} \]

Then

\[ y^{\prime } = \moverset {\infty }{\munderset {n =0}{\sum }}\left (n +r \right ) a_{n} x^{n +r -1} \]

Substituting the above back into the ode gives

\begin{equation} \tag{1} \left (\moverset {\infty }{\munderset {n =0}{\sum }}\left (n +r \right ) a_{n} x^{n +r -1}\right ) x +\left (\moverset {\infty }{\munderset {n =0}{\sum }}a_{n} x^{n +r}\right ) = 0 \end{equation}

Which simplifies to

\begin{equation} \tag{2A} \left (\moverset {\infty }{\munderset {n =0}{\sum }}\left (n +r \right ) a_{n} x^{n +r}\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}a_{n} x^{n +r}\right ) = 0 \end{equation}

The next step is to make all powers of \(x\) be \(n +r\) in each summation term. Going over each summation term above with power of \(x\) in it which is not already \(x^{n +r}\) and adjusting the power and the corresponding index gives Substituting all the above in Eq (2A) gives the following equation where now all powers of \(x\) are the same and equal to \(n +r\).

\begin{equation} \tag{2B} \left (\moverset {\infty }{\munderset {n =0}{\sum }}\left (n +r \right ) a_{n} x^{n +r}\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}a_{n} x^{n +r}\right ) = 0 \end{equation}

The indicial equation is obtained from \(n=0\). From Eq (2) this gives

\[ \left (n +r \right ) a_{n} x^{n +r}+a_{n} x^{n +r} = 0 \]

When \(n=0\) the above becomes

\[ r a_{0} x^{r}+a_{0} x^{r} = 0 \]

The corresponding balance equation is found by replacing \(r\) by \(m\) and \(a\) by \(c\) to avoid confusing terms between particular solution and the homogeneous solution. Hence the balance equation is

\[ \left (x^{m} m +x^{m}\right ) c_{0} = 2 x^{4}+x^{3}+x \]

This equation will used later to find the particular solution.

Since \(a_{0}\neq 0\) then the indicial equation becomes

\[ \left (r +1\right ) x^{r} = 0 \]

Since the above is true for all \(x\) then the indicial equation simplifies to

\[ r +1 = 0 \]

Solving for \(r\) gives the root of the indicial equation as

\[ r=-1 \]

From the above we see that there is no recurrence relation since there is only one summation term. Therefore all \(a_{n}\) terms are zero except for \(a_{0}\). Hence

\begin{align*} y_h &= a_{0} \left (\frac {1}{x}+O\left (x^{6}\right )\right ) \end{align*}

Now the particular solution is found Now we will go over each term in the above, and generate the corresponding particular solution to each term, and at the end add them all.

Looking at the term \(2 x^{4}\), its balance equation is given by

\begin{align*} \left (x^{m} m +x^{m}\right ) c_{0} = 2 x^{4} \end{align*}

Solving the above balance equation gives

\begin{align*} c_{0}&={\frac {2}{5}}\\ m&=4 \end{align*}

Since each sum in the summation equation (2B) start from the same index, then there will be only one term in this particular solution, which is given by

\begin{align*} y_{p_1} &= c_{0} x^{m}\\ &=\frac {2 x^{4}}{5} \end{align*}

Looking at the term \(x^{3}\), its balance equation is given by

\begin{align*} \left (x^{m} m +x^{m}\right ) c_{0} = x^{3} \end{align*}

Solving the above balance equation gives

\begin{align*} c_{0}&={\frac {1}{4}}\\ m&=3 \end{align*}

Since each sum in the summation equation (2B) start from the same index, then there will be only one term in this particular solution, which is given by

\begin{align*} y_{p_2} &= c_{0} x^{m}\\ &=\frac {x^{3}}{4} \end{align*}

Looking at the term \(x\), its balance equation is given by

\begin{align*} \left (x^{m} m +x^{m}\right ) c_{0} = x \end{align*}

Solving the above balance equation gives

\begin{align*} c_{0}&={\frac {1}{2}}\\ m&=1 \end{align*}

Since each sum in the summation equation (2B) start from the same index, then there will be only one term in this particular solution, which is given by

\begin{align*} y_{p_3} &= c_{0} x^{m}\\ &=\frac {x}{2} \end{align*}

Now all the above particular solutions are added which gives

\begin{align*} y_p &= y_{p_{1}}+y_{p_{2}}+y_{p_{3}}\\ &= \frac {2}{5} x^{4}+\frac {1}{4} x^{3}+\frac {1}{2} x \end{align*}

The solution is

\begin{align*} y&=y_h + y_p \\ y &= c_1 \left (\frac {1}{x}+O\left (x^{6}\right )\right )+\frac {2 x^{4}}{5}+\frac {x^{3}}{4}+\frac {x}{2} \\ \end{align*}

Figure 2.90: Slope field \(y^{\prime } x +y = 2 x^{4}+x^{3}+x\)

2.4.7.1 ✓ Maple. Time used: 0.013 (sec). Leaf size: 28

Order:=6; 
ode:=diff(y(x),x)*x+y(x) = 2*x^4+x^3+x; 
dsolve(ode,y(x),type='series',x=0);
 

\[ y = \frac {c_1 \left (1+\operatorname {O}\left (x^{6}\right )\right )}{x}+x \left (\frac {1}{2}+\frac {1}{4} x^{2}+\frac {2}{5} x^{3}+\operatorname {O}\left (x^{5}\right )\right ) \]

Maple trace

Methods for first order ODEs: 
--- Trying classification methods --- 
trying a quadrature 
trying 1st order linear 
<- 1st order linear successful
 

Maple step by step

\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & x \left (\frac {d}{d x}y \left (x \right )\right )+y \left (x \right )=2 x^{4}+x^{3}+x \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 1 \\ {} & {} & \frac {d}{d x}y \left (x \right ) \\ \bullet & {} & \textrm {Isolate the derivative}\hspace {3pt} \\ {} & {} & \frac {d}{d x}y \left (x \right )=-\frac {y \left (x \right )}{x}+2 x^{3}+x^{2}+1 \\ \bullet & {} & \textrm {Group terms with}\hspace {3pt} y \left (x \right )\hspace {3pt}\textrm {on the lhs of the ODE and the rest on the rhs of the ODE}\hspace {3pt} \\ {} & {} & \frac {d}{d x}y \left (x \right )+\frac {y \left (x \right )}{x}=2 x^{3}+x^{2}+1 \\ \bullet & {} & \textrm {The ODE is linear; multiply by an integrating factor}\hspace {3pt} \mu \left (x \right ) \\ {} & {} & \mu \left (x \right ) \left (\frac {d}{d x}y \left (x \right )+\frac {y \left (x \right )}{x}\right )=\mu \left (x \right ) \left (2 x^{3}+x^{2}+1\right ) \\ \bullet & {} & \textrm {Assume the lhs of the ODE is the total derivative}\hspace {3pt} \frac {d}{d x}\left (y \left (x \right ) \mu \left (x \right )\right ) \\ {} & {} & \mu \left (x \right ) \left (\frac {d}{d x}y \left (x \right )+\frac {y \left (x \right )}{x}\right )=\left (\frac {d}{d x}y \left (x \right )\right ) \mu \left (x \right )+y \left (x \right ) \left (\frac {d}{d x}\mu \left (x \right )\right ) \\ \bullet & {} & \textrm {Isolate}\hspace {3pt} \frac {d}{d x}\mu \left (x \right ) \\ {} & {} & \frac {d}{d x}\mu \left (x \right )=\frac {\mu \left (x \right )}{x} \\ \bullet & {} & \textrm {Solve to find the integrating factor}\hspace {3pt} \\ {} & {} & \mu \left (x \right )=x \\ \bullet & {} & \textrm {Integrate both sides with respect to}\hspace {3pt} x \\ {} & {} & \int \left (\frac {d}{d x}\left (y \left (x \right ) \mu \left (x \right )\right )\right )d x =\int \mu \left (x \right ) \left (2 x^{3}+x^{2}+1\right )d x +\mathit {C1} \\ \bullet & {} & \textrm {Evaluate the integral on the lhs}\hspace {3pt} \\ {} & {} & y \left (x \right ) \mu \left (x \right )=\int \mu \left (x \right ) \left (2 x^{3}+x^{2}+1\right )d x +\mathit {C1} \\ \bullet & {} & \textrm {Solve for}\hspace {3pt} y \left (x \right ) \\ {} & {} & y \left (x \right )=\frac {\int \mu \left (x \right ) \left (2 x^{3}+x^{2}+1\right )d x +\mathit {C1}}{\mu \left (x \right )} \\ \bullet & {} & \textrm {Substitute}\hspace {3pt} \mu \left (x \right )=x \\ {} & {} & y \left (x \right )=\frac {\int x \left (2 x^{3}+x^{2}+1\right )d x +\mathit {C1}}{x} \\ \bullet & {} & \textrm {Evaluate the integrals on the rhs}\hspace {3pt} \\ {} & {} & y \left (x \right )=\frac {\frac {2}{5} x^{5}+\frac {1}{4} x^{4}+\frac {1}{2} x^{2}+\mathit {C1}}{x} \\ \bullet & {} & \textrm {Simplify}\hspace {3pt} \\ {} & {} & y \left (x \right )=\frac {8 x^{5}+5 x^{4}+10 x^{2}+20 \mathit {C1}}{20 x} \end {array} \]

2.4.7.2 ✓ Mathematica. Time used: 0.023 (sec). Leaf size: 36

ode=x*D[y[x],x]+y[x]==x+x^3+2*x^4; 
AsymptoticDSolveValue[ode,y[x],{x,0,5}]
 

\[ y(x)\to \frac {\frac {2 x^5}{5}+\frac {x^4}{4}+\frac {x^2}{2}}{x}+\frac {c_1}{x} \]

2.4.7.3 ✗ Sympy

from sympy import * 
x = symbols("x") 
y = Function("y") 
ode = Eq(-2*x**4 - x**3 + x*Derivative(y(x), x) - x + y(x),0) 
ics = {} 
dsolve(ode,func=y(x),ics=ics,hint="1st_power_series",x0=0,n=6)
 

ValueError : ODE -2*x**4 - x**3 + x*Derivative(y(x), x) - x + y(x) does not match hint 1st_power_series