document.write( "Question 1204822: If
\n" ); document.write( "y= A/x + Bx where A and B are constants, show that x²dy/dx +xdy/dx=y \n" ); document.write( "
Algebra.Com's Answer #841355 by Edwin McCravy(20055)\"\" \"About 
You can put this solution on YOUR website!
y= A/x + Bx where A and B are constants, show that x²dy/dx +xdy/dx=y
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document.write( "I don't believe that's true. Let's see if we can find a counter-example.\r\n" );
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document.write( "\"y=A%2Fx%2BBx\"\r\n" );
document.write( "\"y=Ax%5E%28-1%29%2BBx\"\r\n" );
document.write( "\"dy%2Fdx=-Ax%5E%28-2%29%2BB\"\r\n" );
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document.write( "Now if \"x%5E2%2Aexpr%28dy%2Fdx%29+%2Bx%2Aexpr%28dy%2Fdx%29=y\" were true it would be true when A=B=x=1 \r\n" );
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document.write( "\"y=A%2Fx%2BBx=1%2F1%2B1%281%29=1%2B1=2\"\r\n" );
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document.write( "\"dy%2Fdx=-Ax%5E%28-2%29%2BB=-%281%29%281%29%5E%28-2%29%2B1=-1%2B1=0\"\r\n" );
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document.write( "So we substitute in\r\n" );
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document.write( "\"x%5E2%2Aexpr%28dy%2Fdx%29+%2Bx%2Aexpr%28dy%2Fdx%29=y\"\r\n" );
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document.write( "\"%281%29%5E2%2A0%2B1%2A0=2\"\r\n" );
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document.write( "\"0=2\"\r\n" );
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document.write( "False. That's why I don't believe what you are asking to prove is true.\r\n" );
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document.write( "Edwin
 
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