2.1.58 problem 58

Internal problem ID [8446]
Book : Own collection of miscellaneous problems
Section : section 1.0
Problem number : 58
Date solved : Tuesday, December 17, 2024 at 12:51:38 PM
CAS classification : [[_homogeneous, `class C`], _dAlembert]

Solve

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

Solved as first order form A1 ode

Time used: 0.175 (sec)

The given ode has the general form

\begin{align*} y^{\prime } & = B+C f\left ( ax +b y +c\right ) \tag {1} \end{align*}

Comparing (1) to the ode given shows the parameters in the ODE have these values

\begin{align*} B &= -4\\ C &= -4\\ a &= 1\\ b &= -1\\ c &= 0 \end{align*}

This form of ode can be solved by change of variables \(u=ax+b y +c\) which makes the ode separable.

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

Or

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

The ode becomes

\begin{align*} \frac {u' - a}{b} & = B+C f\left ( u\right ) \\ u' & =b B+ b C f\left ( u\right ) +a \\ \frac {du}{b B+b C f\left ( u\right ) +a} &= d x \end{align*}

Integrating gives

\begin{align*} \int \frac {du}{b B+ b C f(u) +a} &=x+c_1\\ \int ^{u}\frac {d\tau }{b B + b C f(\tau ) +a} & = x+c_1 \end{align*}

Replacing back \(u=ax+b y +c\) the above becomes

\begin{equation} \int ^{ax+b y +c}\frac {d\tau }{b B+b C f\left ( \tau \right ) +a} = x+c_{1}\tag {2} \end{equation}

If initial conditions are given as \(y\left ( x_{0}\right ) = y_{0}\), the above becomes

\begin{align*} \int _{0}^{a x_{0}+b y_{0}+c}\frac {d\tau }{b B + b C f\left ( \tau \right ) +a} & =x_{0}+c_{1}\\ c_{1} & =\int _{0}^{ax+by_{0}+c}\frac {d\tau }{b B+ b C f\left ( \tau \right )+a}-x_{0} \end{align*}

Substituting this into (2) gives

\begin{align*} \int ^{ax+by+c}\frac {d\tau }{bB+bC f\left ( \tau \right ) +a}= x+\int _{0}^{ax+by_{0}+c}\frac {d\tau }{bB+bC f\left ( \tau \right ) +a}-x_{0} \tag {3} \end{align*}

Since no initial conditions are given, then using (2) and replacing the values of the parameters into (2) gives the solution as

\[ \int _{}^{-y+x}\frac {1}{5+4 \sin \left (\tau \right )}d \tau = x +c_1 \]

Which simplifies to

\[ \frac {2 \arctan \left (\frac {5 \tan \left (-\frac {y}{2}+\frac {x}{2}\right )}{3}+\frac {4}{3}\right )}{3} = x +c_1 \]

Solving for \(y\) gives

\begin{align*} y &= x -2 \arctan \left (\frac {3 \tan \left (\frac {3 x}{2}+\frac {3 c_1}{2}\right )}{5}-\frac {4}{5}\right ) \\ \end{align*}

Figure 2.140: Slope field plot
\(y^{\prime } = -4 \sin \left (-y+x \right )-4\)

Summary of solutions found

\begin{align*} y &= x -2 \arctan \left (\frac {3 \tan \left (\frac {3 x}{2}+\frac {3 c_1}{2}\right )}{5}-\frac {4}{5}\right ) \\ \end{align*}

Maple step by step solution

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

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 
<- 1st order, canonical coordinates successful 
<- homogeneous successful`
 

Maple dsolve solution

Solving time : 0.048 (sec)
Leaf size : 21

dsolve(diff(y(x),x) = -4*sin(x-y(x))-4, 
       y(x),singsol=all)
 

\[ y = x +2 \arctan \left (\frac {3 \tan \left (-\frac {3 x}{2}+\frac {3 c_{1}}{2}\right )}{5}+\frac {4}{5}\right ) \]

Mathematica DSolve solution

Solving time : 0.0 (sec)
Leaf size : 0

DSolve[{D[y[x],x]==4*Sin[y[x]-x]-4,{}}, 
       y[x],x,IncludeSingularSolutions->True]
 

Timed out