Solving DE with ICs: y''(x) - 8y(x) = 0, y(0)=1, y'(0)=-2

  • Thread starter thepatient
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In summary, the given equation is solved using the substitution method with the variable v and the equation is simplified to v^4 = 16y^2 + c. The initial conditions are used to solve for the constant c. However, there was a mistake made in the integration step which caused confusion. The correct integration should be v^4/4 = 8y^2/2 + c.
  • #1
thepatient
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Given (y')^2 *y'' -8y=0 and ICs y(0)=1 y'(0)= -2

I used substitution dy/dx = v and d^2y/dx^2 = vdv/dy and got to:

v^2*v*v' = 8y
V^3 dv/dy = 8y
v^4/4 =4y + c
V^4 =4y + c

And IC y'(0) = v(0) =-2
So c = 16

v = 2(y^2+1)^(1/4)
dy/dx = 2(y^2+1)^(1/4)

(y^2+1)^(-1/4)dy = 2dx

But here I got stuck. :( did I do the steps wrong? I then tried trig substitution and got:

Sec(u)^3/2 du = 2dx

But couldn't integrate either. This is a DE class so I didn't think it should be like that and maybe I made a mistake. Any help would be greatly appreciated. :]
 
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  • #2
thepatient said:
Given (y')^2 *y'' -8y=0 and ICs y(0)=1 y'(0)= -2

I used substitution dy/dx = v and d^2y/dx^2 = vdv/dy and got to:

v^2*v*v' = 8y
V^3 dv/dy = 8y
v^4/4 =4y + c

The integral of 8y is NOT 4y+ C.

V^4 =4y + c
How did you get this? If you multiply both sides of v^4/4= 4y+ C (which was wrong) by 4 you would get y^4= 16y+ 4c (which you could write as y^4= 16y+ C' since 4 times an unknown constant is still an unknown constant).

And IC y'(0) = v(0) =-2
So c = 16

v = 2(y^2+1)^(1/4)
dy/dx = 2(y^2+1)^(1/4)

(y^2+1)^(-1/4)dy = 2dx

But here I got stuck. :( did I do the steps wrong? I then tried trig substitution and got:

Sec(u)^3/2 du = 2dx

But couldn't integrate either. This is a DE class so I didn't think it should be like that and maybe I made a mistake. Any help would be greatly appreciated. :]
 
  • #3
Oops yes, I had before putting in first IC

V^4 = 16y^2 + C.

I jumped a bit a head of myself while typing this in. XD I just had this on an exam a few hours ago.
 
Last edited:
  • #4
From:
V^3 dv/dy = 8y, I used method of separable variables.

V^3 dv = 8ydy then integrate both sides:

V^4/4 = 8y^2/2 + c
 
  • #5
I accidentally left the squared out on that part. XD it's not easy typing on phone.
 

1. What does DE stand for in this context?

DE stands for "differential equation", which is an equation that involves derivatives of an unknown function.

2. What is the general strategy for solving this type of DE with initial conditions?

The general strategy for solving this type of DE with initial conditions is to first solve the differential equation by finding the general solution, and then use the initial conditions to find the specific solution that satisfies the given conditions.

3. How do you find the general solution for this specific differential equation?

In this case, the differential equation is a second-order homogeneous linear equation, which can be solved by finding the roots of the characteristic equation. In this case, the characteristic equation is r^2 - 8 = 0, which has roots of r = ±√8. Therefore, the general solution is y(x) = c1e^√8x + c2e^-√8x.

4. How do you use the initial conditions to find the specific solution?

To find the specific solution, plug in the values of the initial conditions into the general solution. In this case, we have y(0) = c1e^0 + c2e^0 = c1 + c2 = 1 and y'(0) = c1√8e^0 - c2√8e^0 = c1√8 - c2√8 = -2. This gives us a system of equations that can be solved to find the values of c1 and c2, and thus the specific solution.

5. What is the final solution for this differential equation with the given initial conditions?

The final solution for this differential equation with the given initial conditions is y(x) = (1/√8)e^√8x - (1/√8)e^-√8x.

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