Welcome :: Homework Help and Answers :: Mathskey.com
Welcome to Mathskey.com Question & Answers Community. Ask any math/science homework question and receive answers from other members of the community.

13,435 questions

17,804 answers

1,438 comments

774,932 users

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

0 votes

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

asked Dec 12, 2014 in CALCULUS by anonymous

1 Answer

0 votes

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 = c₁er₁x + c₂er₂x .

So, the general solution of the equation is

y = c₁e- x/2 + c₂ex/2

y(4) = 4 → 4 = c₁e- 4/2 + c₂e4/2 → 4 = c₁e- 2 + c₂e2 ----------> (1)

y = c₁e- x/2 + c₂ex/2

y' = (- 1/2)c₁e- x/2 + (1/2)c₂ex/2

f'(4) = 3 → 3 = (- 1/2)c₁e- 4/2 + (1/2)c₂e4/2 → 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)ex/2 .

y = - e2 + 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 .

answered Dec 12, 2014 by lilly Expert

Related questions

asked May 1, 2014 in PRECALCULUS by anonymous
...