M/32=3/4 multiply both sides by 32
m=96/4
m=24
Answer:
False
Step-by-step explanation:
Like terms have the same variable and are raised to the same exponent, and 6x has a variable of x while 6 has none at all, so they are not like terms.
It rained 3 out of 20 days.
3/20 = 0.15
Move the decimal two points to the right to get the percent.
It rained 15% of the days.
the volume of the box is L x W x H so it should be 7x5x4=140
Given a solution
![y_1(x)=\ln x](https://tex.z-dn.net/?f=y_1%28x%29%3D%5Cln%20x)
, we can attempt to find a solution of the form
![y_2(x)=v(x)y_1(x)](https://tex.z-dn.net/?f=y_2%28x%29%3Dv%28x%29y_1%28x%29)
. We have derivatives
![y_2=v\ln x](https://tex.z-dn.net/?f=y_2%3Dv%5Cln%20x)
![{y_2}'=v'\ln x+\dfrac vx](https://tex.z-dn.net/?f=%7By_2%7D%27%3Dv%27%5Cln%20x%2B%5Cdfrac%20vx)
![{y_2}''=v''\ln x+\dfrac{v'}x+\dfrac{v'x-v}{x^2}=v''\ln x+\dfrac{2v'}x-\dfrac v{x^2}](https://tex.z-dn.net/?f=%7By_2%7D%27%27%3Dv%27%27%5Cln%20x%2B%5Cdfrac%7Bv%27%7Dx%2B%5Cdfrac%7Bv%27x-v%7D%7Bx%5E2%7D%3Dv%27%27%5Cln%20x%2B%5Cdfrac%7B2v%27%7Dx-%5Cdfrac%20v%7Bx%5E2%7D)
Substituting into the ODE, we get
![v''x\ln x+2v'-\dfrac vx+v'\ln x+\dfrac vx=0](https://tex.z-dn.net/?f=v%27%27x%5Cln%20x%2B2v%27-%5Cdfrac%20vx%2Bv%27%5Cln%20x%2B%5Cdfrac%20vx%3D0)
![v''x\ln x+(2+\ln x)v'=0](https://tex.z-dn.net/?f=v%27%27x%5Cln%20x%2B%282%2B%5Cln%20x%29v%27%3D0)
Setting
![w=v'](https://tex.z-dn.net/?f=w%3Dv%27)
, we end up with the linear ODE
![w'x\ln x+(2+\ln x)w=0](https://tex.z-dn.net/?f=w%27x%5Cln%20x%2B%282%2B%5Cln%20x%29w%3D0)
Multiplying both sides by
![\ln x](https://tex.z-dn.net/?f=%5Cln%20x)
, we have
![w' x(\ln x)^2+(2\ln x+(\ln x)^2)w=0](https://tex.z-dn.net/?f=w%27%20x%28%5Cln%20x%29%5E2%2B%282%5Cln%20x%2B%28%5Cln%20x%29%5E2%29w%3D0)
and noting that
![\dfrac{\mathrm d}{\mathrm dx}\left[x(\ln x)^2\right]=(\ln x)^2+\dfrac{2x\ln x}x=(\ln x)^2+2\ln x](https://tex.z-dn.net/?f=%5Cdfrac%7B%5Cmathrm%20d%7D%7B%5Cmathrm%20dx%7D%5Cleft%5Bx%28%5Cln%20x%29%5E2%5Cright%5D%3D%28%5Cln%20x%29%5E2%2B%5Cdfrac%7B2x%5Cln%20x%7Dx%3D%28%5Cln%20x%29%5E2%2B2%5Cln%20x)
we can write the ODE as
![\dfrac{\mathrm d}{\mathrm dx}\left[wx(\ln x)^2\right]=0](https://tex.z-dn.net/?f=%5Cdfrac%7B%5Cmathrm%20d%7D%7B%5Cmathrm%20dx%7D%5Cleft%5Bwx%28%5Cln%20x%29%5E2%5Cright%5D%3D0)
Integrating both sides with respect to
![x](https://tex.z-dn.net/?f=x)
, we get
![wx(\ln x)^2=C_1](https://tex.z-dn.net/?f=wx%28%5Cln%20x%29%5E2%3DC_1)
![w=\dfrac{C_1}{x(\ln x)^2}](https://tex.z-dn.net/?f=w%3D%5Cdfrac%7BC_1%7D%7Bx%28%5Cln%20x%29%5E2%7D)
Now solve for
![v](https://tex.z-dn.net/?f=v)
:
![v'=\dfrac{C_1}{x(\ln x)^2}](https://tex.z-dn.net/?f=v%27%3D%5Cdfrac%7BC_1%7D%7Bx%28%5Cln%20x%29%5E2%7D)
![v=-\dfrac{C_1}{\ln x}+C_2](https://tex.z-dn.net/?f=v%3D-%5Cdfrac%7BC_1%7D%7B%5Cln%20x%7D%2BC_2)
So you have
![y_2=v\ln x=-C_1+C_2\ln x](https://tex.z-dn.net/?f=y_2%3Dv%5Cln%20x%3D-C_1%2BC_2%5Cln%20x)
and given that
![y_1=\ln x](https://tex.z-dn.net/?f=y_1%3D%5Cln%20x)
, the second term in
![y_2](https://tex.z-dn.net/?f=y_2)
is already taken into account in the solution set, which means that
![y_2=1](https://tex.z-dn.net/?f=y_2%3D1)
, i.e. any constant solution is in the solution set.