Differential Equations

Question

3. Solve the following differential equation:
y″+9y=0; y=3, y′=3 at x=Π/3

 

I've never seen a situation with, "y′=3 at x=Π/3"

Answer 1

y" + 9y = 0 is a second order homogeneous linear differential equation with constant coefficients. 

 

To solve this type of differential equation, we first solve the auxiliary equation m² + 9 = 0.                                           This equation has roots m = ±3i = 0±3i.

 

When the auxiliary equation has imaginary roots a±bi, the general solution of the differential equation is

y = Ae^axcos(bx) + Be^axsin(bx).

 

So, the general solution of the given differential equation is y = Ae^0xcos(3x) + Be^0xsin(3x) = Acos(3x) + Bsin(3x)

 

Since y = 3 when x = π/3, we have Acosπ + Bsinπ = 3.  So, -A = 3.  Therefore, A = -3. 

 

So, y = -3cos(3x) + Bsin(3x)

 

     y' = 9sin(3x) +3Bcos(3x)

 

Since y' = 3 when x = π/3, 3 = 9sinπ + 3Bcosπ

 

                                       3 = -3B     B = -1

 

Thus, y = -3cos(3x) - sin(3x)

 

Answer 2

The differential equation itself is the standard equation for simple harmonic motion.   This has the solution:

 

     y = A cos(3 x + Φ)   for some values of the parameters A and Φ.    There are two of these parameters because the differential equation is a second order differential equation.     Values for the parameters can be found by using the data      y(Π/3) =  3   and y'(Π/3) = 3.     Standard trigonometric identities can be used to solve for A and Φ.   This actually easy to carry out since 3 (Π/3) =  Π  and cos(Π) = -1 and sin(Π) =0. 

 

The result is  A = - sqrt(10)  and   Φ =  arctan(-1/3)

 

 

Answer 3

The general solution of a differential equation of the type

 

y'' + k²y = 0

 

is

 

y = Asin(kx) + Bcos(kx),

 

which in this problem is

 

y = Asin(3x) + Bcos(3x)

 

You can verify this by substituting the solution into the differential equation.

 

Now just simply apply the given boundary conditions:

 

y(π/3) = 3 and y'(π/3) = 3

 

y(π/3) = Asin(3·π/3) + Bcos(3·π/3) = 0 - B = 3 ⇒ B = -3

 

So far,

 

y = Asin(3x) - 3cos(3x)

 

y'(π/3) = 3Acos(3·π/3) + 9sin(3·π/3) = -3A + 0 = 3 ⇒ A = -1

 

Final solution:

 

y(x) = -sin(3x) - 3cos(3x)

 

Do not be surprised that the boundary conditions are give at π/3.  They can be given anywhere in the domain of the function, as long as we are given 2 independent equations so we can solve for the 2 unknown coefficients A and B.