2.5.10 problem 10

Internal problem ID [8397]
Book : Own collection of miscellaneous problems
Section : section 5.0
Problem number : 10
Date solved : Sunday, November 10, 2024 at 03:44:24 AM
CAS classification : [_quadrature]

Solve

\begin{align*} y^{\prime }&=\frac {1}{x} \end{align*}

Using series expansion around \(x=0\)

Since this is an inhomogeneous, then let the solution be

\[ y = y_h + y_p \]

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

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

Then

\begin{align*} y^{\prime } &= \moverset {\infty }{\munderset {n =0}{\sum }}\left (n +r \right ) a_{n} x^{n +r -1} \end{align*}

Substituting the above back into the ode gives

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

Which simplifies to

\begin{equation} \tag{2A} \moverset {\infty }{\munderset {n =0}{\sum }}\left (n +r \right ) a_{n} x^{n +r -1} = 0 \end{equation}

The next step is to make all powers of \(x\) be \(n +r -1\) in each summation term. Going over each summation term above with power of \(x\) in it which is not already \(x^{n +r -1}\) 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 \(n +r -1\).

\begin{equation} \tag{2B} \moverset {\infty }{\munderset {n =0}{\sum }}\left (n +r \right ) a_{n} x^{n +r -1} = 0 \end{equation}

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

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

When \(n=0\) the above becomes

\[ r a_{0} x^{-1+r} = 0 \]

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

\begin{align*}m c_{0} x^{-1+m} = \frac {1}{x} \end{align*}

This equation will used later to find the particular solution.

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

\[ r \,x^{-1+r} = 0 \]

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

\[ r = 0 \]

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

\[ r=0 \]

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} x^r \end{align*}

Unable to solve the balance equation \(m c_{0} x^{-1+m} = \frac {1}{x}\) for \(c_{0}\) and \(x\). No particular solution exists.

Unable to find the particular solution. No solution exist.

Maple step by step solution

\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & \frac {d}{d x}y \left (x \right )=\frac {1}{x} \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 1 \\ {} & {} & \frac {d}{d x}y \left (x \right ) \\ \bullet & {} & \textrm {Integrate both sides with respect to}\hspace {3pt} x \\ {} & {} & \int \left (\frac {d}{d x}y \left (x \right )\right )d x =\int \frac {1}{x}d x +\mathit {C1} \\ \bullet & {} & \textrm {Evaluate integral}\hspace {3pt} \\ {} & {} & y \left (x \right )=\ln \left (x \right )+\mathit {C1} \\ \bullet & {} & \textrm {Solve for}\hspace {3pt} y \left (x \right ) \\ {} & {} & y \left (x \right )=\ln \left (x \right )+\mathit {C1} \end {array} \]

Maple trace

`Methods for first order ODEs: 
--- Trying classification methods --- 
trying a quadrature 
<- quadrature successful`
 

Maple dsolve solution

Solving time : 0.034 (sec)
Leaf size : maple_leaf_size

dsolve(diff(y(x),x) = 1/x,y(x), 
       series,x=0)
 

\[ \text {No solution found} \]

Mathematica DSolve solution

Solving time : 0.01 (sec)
Leaf size : 8

AsymptoticDSolveValue[{D[y[x],x]==1/x,{}}, 
       y[x],{x,0,5}]
 

\[ y(x)\to \log (x)+c_1 \]