A Simple Pendulum and its Linearization

Needs["VisualDSolve`"];

\(\(bigpend = 
      PhasePlot[{\(x'\)[t] == y[t], \(y'\)[t] == \(-.1\) y[t] - Sin[x[t]]}, 

        	{x[t], y[t]}, 
	{t, 0, 100}, 
	{x, \(-2\), 7}, 

        	{y, \(-3\), 3}, 
		\ \ \ ShowEquilibria -> True, 
        InitialValues -> {{0, 1}, {0, 2.3}}, 
        PlotStyle -> {AbsoluteThickness[2], Red}, 
\ \ \ 
        ShowEquilibria -> True, 
		VectorField -> True, 
        FieldLogScale -> 9, FieldColor -> Hue, 
		
        FieldThickness -> AbsoluteThickness[2], 
\ \ 	
        AxesLabel -> {"x", "y"}, 
        Ticks -> {Table[k/2\ \ π, {k, \(-1\), 4}], Automatic}
	]; \)\)

\(\(pend = 
      PhasePlot[{\(x'\)[t] == y[t], \(y'\)[t] == \(-.1\) y[t] - Sin[x[t]]}, 

        	{x[t], y[t]}, 
	{t, 0, 100}, 
	{x, \(-π\), π}, 

        	{y, \(-2\), 2}, 
		VectorField -> True, FieldLogScale -> 9, 
        FieldColor -> Red, 
			FieldThickness -> AbsoluteThickness[2], 

        		\ \ ShowEquilibria -> True, 
		
        Ticks -> {Table[k/2\ π, {k, \(-3\), 3}], Automatic}, 
\ \ 	
        AxesLabel -> {"x", "y"}
	]; \)\)

\(\(linpend = 
      PhasePlot[{\(x'\)[t] == y[t], \(y'\)[t] == \(-.1\) y[t] - x[t]}, 

        	{x[t], y[t]}, 
	{t, 0, 100}, 
	{x, \(-π\), π}, 

        	{y, \(-2\), 2}, 
		VectorField -> True, FieldLogScale -> 9, 
        FieldColor -> Blue, 
			FieldThickness -> AbsoluteThickness[2], 

        		\ \ ShowEquilibria -> True, 
\ \ 	
        AxesLabel -> {"x", "y"}, 
			
        Ticks -> {Table[k/2\ π, {k, \(-3\), 3}], Automatic}
	]; \)\)

Show[linpend, pend];

\(\(pend = 
      PhasePlot[{\(x'\)[t] == y[t], \(y'\)[t] == \(-.1\) y[t] - Sin[x[t]]}, 

        	{x[t], y[t]}, 
	{t, 0, 100}, 
	{x, 0, 2  π}, 

        	{y, \(-2\), 2}, 
		VectorField -> True, FieldLogScale -> 9, 
        FieldColor -> Red, 
			FieldThickness -> AbsoluteThickness[2], 

        		\ \ ShowEquilibria -> True, 
\ \ 	
        AxesLabel -> {"x", "y"}, 	
        Ticks -> {Table[k/2\ π, {k, 0, 4}], Automatic}
	]; \)\)

\(\(linpend = 
      PhasePlot[{\(x'\)[t] == y[t], 
          \(y'\)[t] == \(-.1\) y[t] + \((x[t] - π)\)}, 
	{x[t], y[t]}, 
        
	{t, 0, 100}, 
	{x, 0, 2  π}, 
	{y, \(-2\), 2}, 
		
        VectorField -> True, FieldLogScale -> 9, FieldColor -> Blue, 
			
        FieldThickness -> AbsoluteThickness[2], 
		\ \ 
        ShowEquilibria -> True, 
\ \ 	AxesLabel -> {"x", "y"}, 	
        Ticks -> {Table[k/2\ π, {k, 0, 4}], Automatic}
	]; \)\)

Show[linpend, pend];