Run .exe file in Java from file location

I have to open a .exe file from my Java program. So I tried following code First.

Process process = runtime.exec("c:\\program files\\test\\test.exe");

But I was getting some error. Then I found out that the exe has to be launched from that location that is c://program files/test/ only then it will open with out errors. So I decided to write a .bat file and execute so that it will cd to that location and execute the .exe file.

Following is my code:

BufferedWriter fileOut;

String itsFileLocation = "c:\\program files\\test\\"
    System.out.println(itsFileLocation);
    try {
     fileOut = new BufferedWriter(new FileWriter("C:\\test.bat"));
     fileOut.write("cd\\"+"\n");
     fileOut.write("cd "+ itsFileLocation +"\n");
     fileOut.write("test.exe"+"\n");
     fileOut.write("exit"+"\n");
     
     fileOut.close(); // Close the output stream after all output is done.
    } catch (IOException e1) {
     e1.printStackTrace();
    } // Create the Buffered Writer object to write to a file called filename.txt
    Runtime runtime = Runtime.getRuntime();
    try {
     Process process =runtime.exec("cmd /c start C:\\test.bat");
    } catch (IOException e) {
     e.printStackTrace();
    }

The above code works perfectly. However, the command prompt is also opened at the back of my .exe (Application). It closes only after the .exe file exits..

I need to clse my command prompt when my application stats.

My .bat file will be like following after it is written by the program.

cd\
cd C:\Program Files\test\
test.exe
exit