Rearrange the ODE as


Take

, so that

.
Supposing that

, we have

, from which it follows that


So we can write the ODE as

which is linear in

. Multiplying both sides by

, we have

![\dfrac{\mathrm d}{\mathrm dx}\bigg[e^{x^2}u\bigg]=x^3e^{x^2}](https://tex.z-dn.net/?f=%5Cdfrac%7B%5Cmathrm%20d%7D%7B%5Cmathrm%20dx%7D%5Cbigg%5Be%5E%7Bx%5E2%7Du%5Cbigg%5D%3Dx%5E3e%5E%7Bx%5E2%7D)
Integrate both sides with respect to

:
![\displaystyle\int\frac{\mathrm d}{\mathrm dx}\bigg[e^{x^2}u\bigg]\,\mathrm dx=\int x^3e^{x^2}\,\mathrm dx](https://tex.z-dn.net/?f=%5Cdisplaystyle%5Cint%5Cfrac%7B%5Cmathrm%20d%7D%7B%5Cmathrm%20dx%7D%5Cbigg%5Be%5E%7Bx%5E2%7Du%5Cbigg%5D%5C%2C%5Cmathrm%20dx%3D%5Cint%20x%5E3e%5E%7Bx%5E2%7D%5C%2C%5Cmathrm%20dx)

Substitute

, so that

. Then

Integrate the right hand side by parts using



You should end up with



and provided that we restrict

, we can write
Answer:
12
Step-by-step explanation:
7-(-5) = 7+5
7+5 = 12
7-(-5) = 12
"Starting with three, every consecutive line has 2 less than twice the previous line."
this statement means that
your staring line has 3 marbles. You multiply the 3 marblesby 2 so
3x2=6
And then you minus it by 2
6-2=4
which means that you'll get 4 marbles for the next line.
So to get your 6th line, you count how many marbles is on the 5th line but since your diagram doesn't have the 5th line you have to figure out the 5th line by counting how.many marbles is on the 4th line.
4th line = 10 marbles
10×2=20
20-2=18
5th line = 18
18×2=36
36-2=34
So the 6th line has 34 marbles.