Question

In Problems 37-40 solve the given initial-value problem.

y^''+4y=0

y(0)=0, y (ℼ )=0

Short answer

No solution

Step-by-step solution

Determine the initial value of differential equation

The second order differential equation:

y^''+4y=0

Initial conditions

y(0)=0, y (ℼ )=0

Consider y=e^mx as the solution of the differential equation,

Substitute y=e^mx , y^'=me^mx, and y^''=m²me^mxinto y^''+4y=0 to obtain the auxiliary equation.

For any real x, e^mx≠ 0, we have,

m²+4=0

m²= -4

Roots are,

m_1,2=±2i

Find the general solution of this homogeneous equation

General solution of this homogeneous equation y^''+4y=0

y=c₁e^2ix+c₂e^-2ix

= c₁e^o(cos 2x+sin 2x)+c₂e^o(cos 2x-sin 2x)

=k₁ cos 2x+ k₂sin 2x----1

Arbitrary constant:

k₁=c₁+c₂ and k₂=c₁-c₂

Fourth, now to apply the given conditions in the general solution of our given differential equations as the following technique:

We have to apply with the point (x,y)= (0,1) in the general solution (1) as

0=k₁cos (0) + k₂sin (0)

k₁=0

Apply the point (x,y)= (π ,0)

General solution (1)

0=k₁ Cos (π) +k₂ Sin (π)

Apply with point (x,y') =(0,-2)

k₁=0

So, we cannot obtain the value k₂at this equation. So, there is no solution for the given initial problem.

Hence there is no solution for the given initial problem.