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:
C. 210
Step-by-step explanation:
I just took it on edge
The base fee is 4 dollars, and the thing the at changes is x or the hours so the equation is 4 + 2x equals 12. and to solve for x we have to first subtract both sides by 4, getting 2x equals 8 then we divide both sides by 2 to get x equals 4 or 4 hours
<span>Every point went up 4 units and left 3 units.
</span><span>Point X on the triangle moved up 4 and left 3.
That formed a right triangle with legs 4 and 3.
The distance point X moved to point X', d, is the hypotenuse of the right triangle.
</span><span>Using the Pythagoras Theorem
</span><span><span>3^2</span>+<span>4^2</span>=<span>d^2
d = 5
</span></span>
Which relation represents a function? A. {(0, 3), (2, 4), (0, 6)} B. {(-7, 5), (-7, 1), (-10, 3), (-4, 3)} C. {(2, 0), (6, 2), (
xz_007 [3.2K]
If there are two of the same x values in a set, it is not a function. Therefore, the answer is D. Hope this helps, and please mark brainliest!