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
The answer to the question is 1-5
Answer:
60 = x
Step-by-step explanation:
180-144 = 36
36+84 = 120
180-120 = 60
Answer:
0.36878
Step-by-step explanation:
Given that:
Mean number of miles (m) = 2135 miles
Variance = 145924
Sample size (n) = 40
Standard deviation (s) = √variance = √145924 = 382
probability that the mean of a sample of 40 cars would differ from the population mean by less than 29 miles
P( 2135 - 29 < z < 2135 + 29)
Z = (x - m) / s /√n
Z = [(2106 - 2135) / 382 / √40] < z < [(2164 - 2135) / 382 / √40]
Z = (- 29 / 60.399503) < z < (29 / 60.399503)
Z = - 0.4801364 < z < 0.4801364
P(Z < - 0.48) = 0.31561
P(Z < - 0.48) = 0.68439
P(- 0.480 < z < 0.480) = 0.68439 - 0.31561 = 0.36878
= 0.36878
Answer:
y = x - 2
Step-by-step explanation:
You can subtract 4 from the x and y to get where the y-intercept would be.