A function f(x) = e^x*sin 2x. What is f''(x) - f'(x) + f(x)

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justaguide's profile pic

justaguide | College Teacher | (Level 2) Distinguished Educator

Posted on

The function `f(x) = e^x*sin (2x)` .

The first derivative of f(x) can be determined using the product rule.

`f'(x) = (e^x)'*sin (2x) + e^x*(sin (2x))'`

= `e^x*sin (2x) + e^x*2*cos(2x)`

`f''(x) = e^x*sin (2x) + e^x*2*cos(2x) + e^x*2*cos(2x) - 2*2*e^x*sin(2x)`

= `4*e^x*cos(2x) - 3*e^x*sin(2x)`

The sum `f''(x) - f'(x) + f(x)`

= `4*e^x*cos(2x) - 3*e^x*sin(2x) - e^x*sin (2x) - e^x*2*cos(2x) + e^x*sin (2x)`

= `2*e^x*cos(2x) - 3*e^x*sin(2x)`

For the function `f(x) = e^x*sin (2x)` , `f''(x) - f'(x) + f(x) = 2*e^x*cos(2x) - 3*e^x*sin(2x)`

f''(x) - f'(x) + f(x)
oldnick's profile pic

oldnick | (Level 1) Valedictorian

Posted on

`f(x)=e^x sin 2x`

`f'(x)= e^x sin2x+ 2e^xcos2x`

`f''(x)=e^xsin2x+2e^xcos2x+2e^xcos2x-4e^xsin2x=```

 `=e^x(4cos2x-3sin2x)`

`f''(x)-f'(x)+f(x)=e^x(4cos2x-3sin2x-sin2x-2cos2x+sin2x)=`

`=e^x(2cos2x-3sin2x)`

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