[javascript] JavaScript: Class.method vs. Class.prototype.method

For visual learners, when defining the function without .prototype

ExampleClass = function(){};
ExampleClass.method = function(customString){
             console.log((customString !== undefined)? 
                          customString : 
                          "called from func def.");}
ExampleClass.method(); // >> output: `called from func def.`  

var someInstance = new ExampleClass();
someInstance.method('Called from instance');
    // >> error! `someInstance.method is not a function`  

With same code, if .prototype is added,

ExampleClass.prototype.method = function(customString){
             console.log((customString !== undefined)? 
                          customString : 
                          "called from func def.");}
ExampleClass.method();  
      // > error! `ExampleClass.method is not a function.`  

var someInstance = new ExampleClass();
someInstance.method('Called from instance');
                 // > output: `Called from instance`

To make it clearer,

ExampleClass = function(){};
ExampleClass.directM = function(){}  //M for method
ExampleClass.prototype.protoM = function(){}

var instanceOfExample = new ExampleClass();

ExampleClass.directM();     ? works
instanceOfExample.directM();   x Error!

ExampleClass.protoM();     x Error!
instanceOfExample.protoM();  ? works

****Note for the example above, someInstance.method() won't be executed as,
ExampleClass.method() causes error & execution cannot continue.
But for the sake of illustration & easy understanding, I've kept this sequence.****

Results generated from chrome developer console & JS Bin
Click on the jsbin link above to step through the code.
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