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Determine if the given (y) is a solution to the differential equation

y'''-y''+y'= cos x , y= sin x

User Mossawi
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Final answer:

After calculating the derivatives of y = sin x and substituting them into the differential equation y''' - y'' + y' = cos x, we find that the left side of the equation simplifies to cos x, which is equal to the right side. Thus, y = sin x is a solution to the differential equation.

Step-by-step explanation:

We are tasked with determining if the function y = sin x is a solution to the differential equation y''' - y'' + y' = cos x. First, we calculate the first, second, and third derivatives of y = sin x.

  • The first derivative (y') is cos x.
  • The second derivative (y'') is -sin x.
  • The third derivative (y''') is -cos x.

We then substitute these into the given differential equation:

(-cos x) - (-sin x) + (cos x) = cos x

Simplifying, we get:

cos x = cos x

Since the left side of the equation equals the right side, y = sin x is indeed a solution to the differential equation.

User Faigjaz
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