Substitute

, so that

![\dfrac{\mathrm d^2y}{\mathrm dx^2}=\dfrac{\mathrm d}{\mathrm dx}\left[\dfrac1x\dfrac{\mathrm dy}{\mathrm dz}\right]=-\dfrac1{x^2}\dfrac{\mathrm dy}{\mathrm dz}+\dfrac1x\left(\dfrac1x\dfrac{\mathrm d^2y}{\mathrm dz^2}\right)=\dfrac1{x^2}\left(\dfrac{\mathrm d^2y}{\mathrm dz^2}-\dfrac{\mathrm dy}{\mathrm dz}\right)](https://tex.z-dn.net/?f=%5Cdfrac%7B%5Cmathrm%20d%5E2y%7D%7B%5Cmathrm%20dx%5E2%7D%3D%5Cdfrac%7B%5Cmathrm%20d%7D%7B%5Cmathrm%20dx%7D%5Cleft%5B%5Cdfrac1x%5Cdfrac%7B%5Cmathrm%20dy%7D%7B%5Cmathrm%20dz%7D%5Cright%5D%3D-%5Cdfrac1%7Bx%5E2%7D%5Cdfrac%7B%5Cmathrm%20dy%7D%7B%5Cmathrm%20dz%7D%2B%5Cdfrac1x%5Cleft%28%5Cdfrac1x%5Cdfrac%7B%5Cmathrm%20d%5E2y%7D%7B%5Cmathrm%20dz%5E2%7D%5Cright%29%3D%5Cdfrac1%7Bx%5E2%7D%5Cleft%28%5Cdfrac%7B%5Cmathrm%20d%5E2y%7D%7B%5Cmathrm%20dz%5E2%7D-%5Cdfrac%7B%5Cmathrm%20dy%7D%7B%5Cmathrm%20dz%7D%5Cright%29)
Then the ODE becomes


which has the characteristic equation

with roots at

. This means the characteristic solution for

is

and in terms of

, this is

From the given initial conditions, we find


so the particular solution to the IVP is
Answer:
744 in³
Step-by-step explanation:
Since you are filling the larger box with both rectangular prism and Styrofoam peanuts, you need to find the overall volume of the larger box and subtract the volume of the glass box to find the amount of space that the Styrofoam peanuts need to take up.
Volume (prism) = Bh, where B = area of the base, h = height
Larger Box: V = 10 x 10 x 15 = 1500 in³
Glass Box: V = 7 x 9 x 12 = 756 in³
1500 - 756 = 744 in³ of Styrofoam peanuts
Answer: the answer is x+8= 1
Answer:
Decay
Step-by-step explanation:
The money is being lost, so it is decaying
The answer is fill its tanks with air so it floats to the top, but I will solve for the distance since I am nice
so we do
decent means negative
cimb=positive
add
decends 45 and 1/2
-45 and 1/2
climbs 39
+39
decedns 18 and 3/4
-18 and 3/4
add
-45 and 1/2+39-18 and 3/4
add
convert 1/2 to 2/4
gropu negatives
-45 and 2/4-18 and 3/4+39
add
-63 and 5/4+39
-64 and 1/4+39
-25 and 1/4
that is how far dow the submarine is
so the submarine needs to gu 25 and 1/4 ft up to reach the csurface