2.37 problem Problem 18(b)

Internal problem ID [12257]
Internal file name [OUTPUT/10910_Thursday_September_28_2023_01_08_30_AM_54738996/index.tex]

Book: APPLIED DIFFERENTIAL EQUATIONS The Primary Course by Vladimir A. Dobrushkin. CRC Press 2015
Section: Chapter 4, Second and Higher Order Linear Differential Equations. Problems page 221
Problem number: Problem 18(b).
ODE order: 2.
ODE degree: 1.

The type(s) of ODE detected by this program : "exact linear second order ode", "second_order_integrable_as_is"

Maple gives the following as the ode type

[[_2nd_order, _exact, _linear, _homogeneous]]

\[ \boxed {x y^{\prime \prime }+\sin \left (x \right ) y^{\prime }+\cos \left (x \right ) y=0} \]

2.37.1 Solving as second order integrable as is ode

Integrating both sides of the ODE w.r.t \(x\) gives \begin {align*} \int \left (x y^{\prime \prime }+\sin \left (x \right ) y^{\prime }+\cos \left (x \right ) y\right )d x &= 0 \\ \left (\sin \left (x \right )-1\right ) y+x y^{\prime } = c_{1} \end {align*}

Which is now solved for \(y\).

Entering Linear first order ODE solver. In canonical form a linear first order is \begin {align*} y^{\prime } + p(x)y &= q(x) \end {align*}

Where here \begin {align*} p(x) &=\frac {\sin \left (x \right )-1}{x}\\ q(x) &=\frac {c_{1}}{x} \end {align*}

Hence the ode is \begin {align*} y^{\prime }+\frac {\left (\sin \left (x \right )-1\right ) y}{x} = \frac {c_{1}}{x} \end {align*}

The integrating factor \(\mu \) is \[ \mu = {\mathrm e}^{\int \frac {\sin \left (x \right )-1}{x}d x} \] The ode becomes \begin {align*} \frac {\mathop {\mathrm {d}}}{ \mathop {\mathrm {d}x}}\left ( \mu y\right ) &= \left (\mu \right ) \left (\frac {c_{1}}{x}\right ) \\ \frac {\mathop {\mathrm {d}}}{ \mathop {\mathrm {d}x}} \left ({\mathrm e}^{\int \frac {\sin \left (x \right )-1}{x}d x} y\right ) &= \left ({\mathrm e}^{\int \frac {\sin \left (x \right )-1}{x}d x}\right ) \left (\frac {c_{1}}{x}\right )\\ \mathrm {d} \left ({\mathrm e}^{\int \frac {\sin \left (x \right )-1}{x}d x} y\right ) &= \left (\frac {c_{1} {\mathrm e}^{\int \frac {\sin \left (x \right )-1}{x}d x}}{x}\right )\, \mathrm {d} x \end {align*}

Integrating gives \begin {align*} {\mathrm e}^{\int \frac {\sin \left (x \right )-1}{x}d x} y &= \int {\frac {c_{1} {\mathrm e}^{\int \frac {\sin \left (x \right )-1}{x}d x}}{x}\,\mathrm {d} x}\\ {\mathrm e}^{\int \frac {\sin \left (x \right )-1}{x}d x} y &= \int \frac {c_{1} {\mathrm e}^{\int \frac {\sin \left (x \right )-1}{x}d x}}{x}d x + c_{2} \end {align*}

Dividing both sides by the integrating factor \(\mu ={\mathrm e}^{\int \frac {\sin \left (x \right )-1}{x}d x}\) results in \begin {align*} y &= {\mathrm e}^{-\left (\int \frac {\sin \left (x \right )-1}{x}d x \right )} \left (\int \frac {c_{1} {\mathrm e}^{\int \frac {\sin \left (x \right )-1}{x}d x}}{x}d x \right )+c_{2} {\mathrm e}^{-\left (\int \frac {\sin \left (x \right )-1}{x}d x \right )} \end {align*}

which simplifies to \begin {align*} y &= {\mathrm e}^{-\left (\int \frac {\sin \left (x \right )-1}{x}d x \right )} \left (c_{1} \left (\int \frac {{\mathrm e}^{\int \frac {\sin \left (x \right )-1}{x}d x}}{x}d x \right )+c_{2} \right ) \end {align*}

Summary

The solution(s) found are the following \begin{align*} \tag{1} y &= {\mathrm e}^{-\left (\int \frac {\sin \left (x \right )-1}{x}d x \right )} \left (c_{1} \left (\int \frac {{\mathrm e}^{\int \frac {\sin \left (x \right )-1}{x}d x}}{x}d x \right )+c_{2} \right ) \\ \end{align*}

Verification of solutions

\[ y = {\mathrm e}^{-\left (\int \frac {\sin \left (x \right )-1}{x}d x \right )} \left (c_{1} \left (\int \frac {{\mathrm e}^{\int \frac {\sin \left (x \right )-1}{x}d x}}{x}d x \right )+c_{2} \right ) \] Verified OK.

2.37.2 Solving as type second_order_integrable_as_is (not using ABC version)

Writing the ode as \[ x y^{\prime \prime }+\sin \left (x \right ) y^{\prime }+\cos \left (x \right ) y = 0 \] Integrating both sides of the ODE w.r.t \(x\) gives \begin {align*} \int \left (x y^{\prime \prime }+\sin \left (x \right ) y^{\prime }+\cos \left (x \right ) y\right )d x &= 0 \\ \left (\sin \left (x \right )-1\right ) y+x y^{\prime } = c_{1} \end {align*}

Which is now solved for \(y\).

Entering Linear first order ODE solver. In canonical form a linear first order is \begin {align*} y^{\prime } + p(x)y &= q(x) \end {align*}

Where here \begin {align*} p(x) &=\frac {\sin \left (x \right )-1}{x}\\ q(x) &=\frac {c_{1}}{x} \end {align*}

Hence the ode is \begin {align*} y^{\prime }+\frac {\left (\sin \left (x \right )-1\right ) y}{x} = \frac {c_{1}}{x} \end {align*}

The integrating factor \(\mu \) is \[ \mu = {\mathrm e}^{\int \frac {\sin \left (x \right )-1}{x}d x} \] The ode becomes \begin {align*} \frac {\mathop {\mathrm {d}}}{ \mathop {\mathrm {d}x}}\left ( \mu y\right ) &= \left (\mu \right ) \left (\frac {c_{1}}{x}\right ) \\ \frac {\mathop {\mathrm {d}}}{ \mathop {\mathrm {d}x}} \left ({\mathrm e}^{\int \frac {\sin \left (x \right )-1}{x}d x} y\right ) &= \left ({\mathrm e}^{\int \frac {\sin \left (x \right )-1}{x}d x}\right ) \left (\frac {c_{1}}{x}\right )\\ \mathrm {d} \left ({\mathrm e}^{\int \frac {\sin \left (x \right )-1}{x}d x} y\right ) &= \left (\frac {c_{1} {\mathrm e}^{\int \frac {\sin \left (x \right )-1}{x}d x}}{x}\right )\, \mathrm {d} x \end {align*}

Integrating gives \begin {align*} {\mathrm e}^{\int \frac {\sin \left (x \right )-1}{x}d x} y &= \int {\frac {c_{1} {\mathrm e}^{\int \frac {\sin \left (x \right )-1}{x}d x}}{x}\,\mathrm {d} x}\\ {\mathrm e}^{\int \frac {\sin \left (x \right )-1}{x}d x} y &= \int \frac {c_{1} {\mathrm e}^{\int \frac {\sin \left (x \right )-1}{x}d x}}{x}d x + c_{2} \end {align*}

Dividing both sides by the integrating factor \(\mu ={\mathrm e}^{\int \frac {\sin \left (x \right )-1}{x}d x}\) results in \begin {align*} y &= {\mathrm e}^{-\left (\int \frac {\sin \left (x \right )-1}{x}d x \right )} \left (\int \frac {c_{1} {\mathrm e}^{\int \frac {\sin \left (x \right )-1}{x}d x}}{x}d x \right )+c_{2} {\mathrm e}^{-\left (\int \frac {\sin \left (x \right )-1}{x}d x \right )} \end {align*}

which simplifies to \begin {align*} y &= {\mathrm e}^{-\left (\int \frac {\sin \left (x \right )-1}{x}d x \right )} \left (c_{1} \left (\int \frac {{\mathrm e}^{\int \frac {\sin \left (x \right )-1}{x}d x}}{x}d x \right )+c_{2} \right ) \end {align*}

Summary

The solution(s) found are the following \begin{align*} \tag{1} y &= {\mathrm e}^{-\left (\int \frac {\sin \left (x \right )-1}{x}d x \right )} \left (c_{1} \left (\int \frac {{\mathrm e}^{\int \frac {\sin \left (x \right )-1}{x}d x}}{x}d x \right )+c_{2} \right ) \\ \end{align*}

Verification of solutions

\[ y = {\mathrm e}^{-\left (\int \frac {\sin \left (x \right )-1}{x}d x \right )} \left (c_{1} \left (\int \frac {{\mathrm e}^{\int \frac {\sin \left (x \right )-1}{x}d x}}{x}d x \right )+c_{2} \right ) \] Verified OK.

2.37.3 Solving as exact linear second order ode ode

An ode of the form \begin {align*} p \left (x \right ) y^{\prime \prime }+q \left (x \right ) y^{\prime }+r \left (x \right ) y&=s \left (x \right ) \end {align*}

is exact if \begin {align*} p''(x) - q'(x) + r(x) &= 0 \tag {1} \end {align*}

For the given ode we have \begin {align*} p(x) &= x\\ q(x) &= \sin \left (x \right )\\ r(x) &= \cos \left (x \right )\\ s(x) &= 0 \end {align*}

Hence \begin {align*} p''(x) &= 0\\ q'(x) &= \cos \left (x \right ) \end {align*}

Therefore (1) becomes \begin {align*} 0- \left (\cos \left (x \right )\right ) + \left (\cos \left (x \right )\right )&=0 \end {align*}

Hence the ode is exact. Since we now know the ode is exact, it can be written as \begin {align*} \left (p \left (x \right ) y^{\prime }+\left (q \left (x \right )-p^{\prime }\left (x \right )\right ) y\right )' &= s(x) \end {align*}

Integrating gives \begin {align*} p \left (x \right ) y^{\prime }+\left (q \left (x \right )-p^{\prime }\left (x \right )\right ) y&=\int {s \left (x \right )\, dx} \end {align*}

Substituting the above values for \(p,q,r,s\) gives \begin {align*} \left (\sin \left (x \right )-1\right ) y+x y^{\prime }&=c_{1} \end {align*}

We now have a first order ode to solve which is \begin {align*} \left (\sin \left (x \right )-1\right ) y+x y^{\prime } = c_{1} \end {align*}

Entering Linear first order ODE solver. In canonical form a linear first order is \begin {align*} y^{\prime } + p(x)y &= q(x) \end {align*}

Where here \begin {align*} p(x) &=\frac {\sin \left (x \right )-1}{x}\\ q(x) &=\frac {c_{1}}{x} \end {align*}

Hence the ode is \begin {align*} y^{\prime }+\frac {\left (\sin \left (x \right )-1\right ) y}{x} = \frac {c_{1}}{x} \end {align*}

The integrating factor \(\mu \) is \[ \mu = {\mathrm e}^{\int \frac {\sin \left (x \right )-1}{x}d x} \] The ode becomes \begin {align*} \frac {\mathop {\mathrm {d}}}{ \mathop {\mathrm {d}x}}\left ( \mu y\right ) &= \left (\mu \right ) \left (\frac {c_{1}}{x}\right ) \\ \frac {\mathop {\mathrm {d}}}{ \mathop {\mathrm {d}x}} \left ({\mathrm e}^{\int \frac {\sin \left (x \right )-1}{x}d x} y\right ) &= \left ({\mathrm e}^{\int \frac {\sin \left (x \right )-1}{x}d x}\right ) \left (\frac {c_{1}}{x}\right )\\ \mathrm {d} \left ({\mathrm e}^{\int \frac {\sin \left (x \right )-1}{x}d x} y\right ) &= \left (\frac {c_{1} {\mathrm e}^{\int \frac {\sin \left (x \right )-1}{x}d x}}{x}\right )\, \mathrm {d} x \end {align*}

Integrating gives \begin {align*} {\mathrm e}^{\int \frac {\sin \left (x \right )-1}{x}d x} y &= \int {\frac {c_{1} {\mathrm e}^{\int \frac {\sin \left (x \right )-1}{x}d x}}{x}\,\mathrm {d} x}\\ {\mathrm e}^{\int \frac {\sin \left (x \right )-1}{x}d x} y &= \int \frac {c_{1} {\mathrm e}^{\int \frac {\sin \left (x \right )-1}{x}d x}}{x}d x + c_{2} \end {align*}

Dividing both sides by the integrating factor \(\mu ={\mathrm e}^{\int \frac {\sin \left (x \right )-1}{x}d x}\) results in \begin {align*} y &= {\mathrm e}^{-\left (\int \frac {\sin \left (x \right )-1}{x}d x \right )} \left (\int \frac {c_{1} {\mathrm e}^{\int \frac {\sin \left (x \right )-1}{x}d x}}{x}d x \right )+c_{2} {\mathrm e}^{-\left (\int \frac {\sin \left (x \right )-1}{x}d x \right )} \end {align*}

which simplifies to \begin {align*} y &= {\mathrm e}^{-\left (\int \frac {\sin \left (x \right )-1}{x}d x \right )} \left (c_{1} \left (\int \frac {{\mathrm e}^{\int \frac {\sin \left (x \right )-1}{x}d x}}{x}d x \right )+c_{2} \right ) \end {align*}

Summary

The solution(s) found are the following \begin{align*} \tag{1} y &= {\mathrm e}^{-\left (\int \frac {\sin \left (x \right )-1}{x}d x \right )} \left (c_{1} \left (\int \frac {{\mathrm e}^{\int \frac {\sin \left (x \right )-1}{x}d x}}{x}d x \right )+c_{2} \right ) \\ \end{align*}

Verification of solutions

\[ y = {\mathrm e}^{-\left (\int \frac {\sin \left (x \right )-1}{x}d x \right )} \left (c_{1} \left (\int \frac {{\mathrm e}^{\int \frac {\sin \left (x \right )-1}{x}d x}}{x}d x \right )+c_{2} \right ) \] Verified OK.

Maple trace

`Methods for second order ODEs: 
--- Trying classification methods --- 
trying a symmetry of the form [xi=0, eta=F(x)] 
   One independent solution has integrals. Trying a hypergeometric solution free of integrals... 
   -> hyper3: Equivalence to 2F1, 1F1 or 0F1 under a power @ Moebius 
No hypergeometric solution was found. 
<- linear_1 successful`
 

✓ Solution by Maple

Time used: 0.016 (sec). Leaf size: 24

dsolve(x*diff(y(x),x$2)+sin(x)*diff(y(x),x)+cos(x)*y(x)=0,y(x), singsol=all)
 

\[ y \left (x \right ) = \left (c_{1} \left (\int \frac {{\mathrm e}^{\operatorname {Si}\left (x \right )}}{x^{2}}d x \right )+c_{2} \right ) {\mathrm e}^{-\operatorname {Si}\left (x \right )} x \]

✗ Solution by Mathematica

Time used: 0.0 (sec). Leaf size: 0

DSolve[x*y''[x]+Sin[x]*y'[x]+Cos[x]*y[x]==0,y[x],x,IncludeSingularSolutions -> True]
 

Not solved