true or false?

Question

y=sinx+cosx satisfies the differential equation ( (d²y)/(dt²) )+y=0

 

true or false?

Answer 1

If y = sinx + cosx , then dy/dx = cosx - sinx

 

                               &   d²y/dx² = -sinx - cosx

 

So, d²y/dx² + y = (-sinx - cosx) + (sinx + cosx) = 0

 

Therefore, y = sinx + cosx is a solution of the given differential equation

Answer 2

Lets see.

 

 

Let   y = e^rx

       y' = re^rx

       y'' = r²e^rx

 

 

r²e^rx + e^rx = 0

 

e^rx(r² + 1) = 0

 

 

r² + 1 = 0

 

r² = -1

 

r = ±√-1

 

r = 0 ± i

 

r = a ± bi

 

 

The general form of the solution is

 

y = e^axcosbx + e^axsinbx

 

 

where a is real number part in the solution r and b is the coefficient of the complex number part in the solution r. 

 

 

The value of your given y satisfies the equation if we take the positive conjugate of r.