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
1/2 gallon of milk makes 10 glasses.

x

=
Answer:
-2.86 , -0.37 , 0.19 , 0.91 , 1.46
Step-by-step explanation:
Answer:
Step-by-step explanation:
Given that a parking lot has two entrances. Cars arrive at entrance I according to a Poisson distribution at an average of 3 per hour and at entrance II according to a Poisson distribution at an average of 2 per hour.
Assuming the number of cars arriving at the two parking lots are independent we have total number of cars arriving X is Poisson with parameter 3+2 = 5
X is Poisson with mean = 5
the probability that a total of 3 cars will arrive at the parking lot in a given hour
= P(X=3) = 0.1404
b) the probability that less than 3 cars will arrive at the parking lot in a given hour
= P(X<3)
= P(0)+P(1)+P(2)
= 0.1247
Answer:
58.1 and 17
Step-by-step explanation:
For the first triangles the similarity ratio is 1:7 so x is 8.3 × 7 = 58.1
For the second triangles the similarity ratio is 1:5 so x is 3.4 × 5 = 17