Solve the initial value problem: d^2y/dx^2 = =y/4?

Question

f'(4)=3 ,y(4)=4..

Answer 1

From the given data : y'' = y/4.

4y'' = y

4y'' - y = 0.

The above differential equation is in the form of ay'' + by' + c = 0, a, b, and c are constants, but a ≠ 0.

If the roots r₁ and r₂ of the auxiliary equation ar² + br + c = 0 are found by using the quadratic formula.

Compare the equation 4y'' - y = 0 with quadratic equation ar² + br + c = 0.

a = 4, b = 0, and c = - 1.

r₁ = [- b - √(b² - 4ac)] / 2a  and r₂ = [- b + √(b² - 4ac)] / 2a

The roots : r₁ = [- 0 - √(0² - 4*4*(- 1))] / 2*4  and r₂ =[- 0 + √(0² - 4*4*(- 1))] / 2*4

r₁ = - √16 / 8  and r₂ = √16 / 8

r₁ = - 4 / 8  and r₂ = 4 / 8

r₁ = - 1 / 2  and r₂ = 1 / 2.

The roots r₁ and r₂ auxiliary equation : ar² + br + c = 0 are real and unequal , then the general solution of ay'' + by' + c = 0 is

y = c₁e^r₁x + c₂e^r₂x .

So, the general solution of the equation is

y = c₁e^{- x/2} + c₂e^x/2

y(4) = 4 → 4 = c₁e^{- 4/2} + c₂e^4/2 → 4 = c₁e^{- 2} + c₂e² ----------> (1)

y = c₁e^{- x/2} + c₂e^x/2

y' = (- 1/2)c₁e^{- x/2} + (1/2)c₂e^x/2

f'(4) = 3 → 3 = (- 1/2)c₁e^{- 4/2} + (1/2)c₂e^4/2 → 6 = - c₁e^{- 2} + c₂e² ----------> (2)

Solve equation (1) and (2).

c₁e^{- 2} + c₂e² = 4

- c₁e^{- 2} + c₂e² = 6

(+)_____________

2c₂e² = 10

c₂e² = 5

c₂ = 5/e² .

Substitute c₂ = 5/e² in equation 1 : c₁e^{- 2} + c₂e² = 4.

c₁e^{- 2} + (5/e²)e² = 4

c₁e^{- 2} + 5 = 4

c₁e^{- 2} = 4 - 5 = - 1

c₁ = - 1/e^{- 2} = - e² .

The general solution is y = (- e²)e^{- x/2} + (5e^{- 2})e^x/2 .

y = - e^{2 + x/2} + 5e^{- 2 + x/2}

y =  5e^{(x - 4)/2} - e^{(x + 4)/2} .

Therefore, y =  5e^{(x - 4)/2} - e^{(x + 4)/2} .