Answer:
The value of X is 6. X=6
Step-by-step explanation:
Given a solution

, we can attempt to find a solution of the form

. We have derivatives



Substituting into the ODE, we get


Setting

, we end up with the linear ODE

Multiplying both sides by

, we have

and noting that
![\dfrac{\mathrm d}{\mathrm dx}\left[x(\ln x)^2\right]=(\ln x)^2+\dfrac{2x\ln x}x=(\ln x)^2+2\ln x](https://tex.z-dn.net/?f=%5Cdfrac%7B%5Cmathrm%20d%7D%7B%5Cmathrm%20dx%7D%5Cleft%5Bx%28%5Cln%20x%29%5E2%5Cright%5D%3D%28%5Cln%20x%29%5E2%2B%5Cdfrac%7B2x%5Cln%20x%7Dx%3D%28%5Cln%20x%29%5E2%2B2%5Cln%20x)
we can write the ODE as
![\dfrac{\mathrm d}{\mathrm dx}\left[wx(\ln x)^2\right]=0](https://tex.z-dn.net/?f=%5Cdfrac%7B%5Cmathrm%20d%7D%7B%5Cmathrm%20dx%7D%5Cleft%5Bwx%28%5Cln%20x%29%5E2%5Cright%5D%3D0)
Integrating both sides with respect to

, we get


Now solve for

:


So you have

and given that

, the second term in

is already taken into account in the solution set, which means that

, i.e. any constant solution is in the solution set.
Answer:
Step-by-step explanation:
A man steps out of a plane at 4,000m of height above the ground.The point at which he jumps out of the plane would make a good reference point. However, if his acceleration is going to change as a result of him opening his parachute 2000m above the ground, a good reference point would be there. Keep in mind though, that his velocity at that instant would need to be known for it to be useful- otherwise the airplane reference point would be just as good with appropriate modeling....
Answer:
12
Step-by-step explanation:
Answer:
y - 4 = (1/3)(x - 9) or x - 3y = -3
Step-by-step explanation:
Here we're told that the slope is 1/3 and that (9, 4) is a point on the line. Use the point-slope formula y - k = m(x - h). Substitute 4 for k, 9 for h and 1/3 for m:
y - 4 = (1/3)(x - 9) (point-slope formula)
3y - 12 = x - 9 (standard form of the same equation) = >
-12 + 9 = x - 3y, or x - 3y = -3 (standard form)