Answer:
4.5
Step-by-step explanation:
1±4±2±8±4±8=27
27÷6=4.5
The first equation is linear:
![x\dfrac{\mathrm dy}{\mathrm dx}-y=x^2\sin x](https://tex.z-dn.net/?f=x%5Cdfrac%7B%5Cmathrm%20dy%7D%7B%5Cmathrm%20dx%7D-y%3Dx%5E2%5Csin%20x)
Divide through by
![x^2](https://tex.z-dn.net/?f=x%5E2)
to get
![\dfrac1x\dfrac{\mathrm dy}{\mathrm dx}-\dfrac1{x^2}y=\sin x](https://tex.z-dn.net/?f=%5Cdfrac1x%5Cdfrac%7B%5Cmathrm%20dy%7D%7B%5Cmathrm%20dx%7D-%5Cdfrac1%7Bx%5E2%7Dy%3D%5Csin%20x)
and notice that the left hand side can be consolidated as a derivative of a product. After doing so, you can integrate both sides and solve for
![y](https://tex.z-dn.net/?f=y)
.
![\dfrac{\mathrm d}{\mathrm dx}\left[\dfrac1xy\right]=\sin x](https://tex.z-dn.net/?f=%5Cdfrac%7B%5Cmathrm%20d%7D%7B%5Cmathrm%20dx%7D%5Cleft%5B%5Cdfrac1xy%5Cright%5D%3D%5Csin%20x)
![\implies\dfrac1xy=\displaystyle\int\sin x\,\mathrm dx=-\cos x+C](https://tex.z-dn.net/?f=%5Cimplies%5Cdfrac1xy%3D%5Cdisplaystyle%5Cint%5Csin%20x%5C%2C%5Cmathrm%20dx%3D-%5Ccos%20x%2BC)
![\implies y=-x\cos x+Cx](https://tex.z-dn.net/?f=%5Cimplies%20y%3D-x%5Ccos%20x%2BCx)
- - -
The second equation is also linear:
![x^2y'+x(x+2)y=e^x](https://tex.z-dn.net/?f=x%5E2y%27%2Bx%28x%2B2%29y%3De%5Ex)
Multiply both sides by
![e^x](https://tex.z-dn.net/?f=e%5Ex)
to get
![x^2e^xy'+x(x+2)e^xy=e^{2x}](https://tex.z-dn.net/?f=x%5E2e%5Exy%27%2Bx%28x%2B2%29e%5Exy%3De%5E%7B2x%7D)
and recall that
![(x^2e^x)'=2xe^x+x^2e^x=x(x+2)e^x](https://tex.z-dn.net/?f=%28x%5E2e%5Ex%29%27%3D2xe%5Ex%2Bx%5E2e%5Ex%3Dx%28x%2B2%29e%5Ex)
, so we can write
![(x^2e^xy)'=e^{2x}](https://tex.z-dn.net/?f=%28x%5E2e%5Exy%29%27%3De%5E%7B2x%7D)
![\implies x^2e^xy=\displaystyle\int e^{2x}\,\mathrm dx=\frac12e^{2x}+C](https://tex.z-dn.net/?f=%5Cimplies%20x%5E2e%5Exy%3D%5Cdisplaystyle%5Cint%20e%5E%7B2x%7D%5C%2C%5Cmathrm%20dx%3D%5Cfrac12e%5E%7B2x%7D%2BC)
![\implies y=\dfrac{e^x}{2x^2}+\dfrac C{x^2e^x}](https://tex.z-dn.net/?f=%5Cimplies%20y%3D%5Cdfrac%7Be%5Ex%7D%7B2x%5E2%7D%2B%5Cdfrac%20C%7Bx%5E2e%5Ex%7D)
- - -
Yet another linear ODE:
![\cos x\dfrac{\mathrm dy}{\mathrm dx}+\sin x\,y=1](https://tex.z-dn.net/?f=%5Ccos%20x%5Cdfrac%7B%5Cmathrm%20dy%7D%7B%5Cmathrm%20dx%7D%2B%5Csin%20x%5C%2Cy%3D1)
Divide through by
![\cos^2x](https://tex.z-dn.net/?f=%5Ccos%5E2x)
, giving
![\dfrac1{\cos x}\dfrac{\mathrm dy}{\mathrm dx}+\dfrac{\sin x}{\cos^2x}y=\dfrac1{\cos^2x}](https://tex.z-dn.net/?f=%5Cdfrac1%7B%5Ccos%20x%7D%5Cdfrac%7B%5Cmathrm%20dy%7D%7B%5Cmathrm%20dx%7D%2B%5Cdfrac%7B%5Csin%20x%7D%7B%5Ccos%5E2x%7Dy%3D%5Cdfrac1%7B%5Ccos%5E2x%7D)
![\sec x\dfrac{\mathrm dy}{\mathrm dx}+\sec x\tan x\,y=\sec^2x](https://tex.z-dn.net/?f=%5Csec%20x%5Cdfrac%7B%5Cmathrm%20dy%7D%7B%5Cmathrm%20dx%7D%2B%5Csec%20x%5Ctan%20x%5C%2Cy%3D%5Csec%5E2x)
![\dfrac{\mathrm d}{\mathrm dx}[\sec x\,y]=\sec^2x](https://tex.z-dn.net/?f=%5Cdfrac%7B%5Cmathrm%20d%7D%7B%5Cmathrm%20dx%7D%5B%5Csec%20x%5C%2Cy%5D%3D%5Csec%5E2x)
![\implies\sec x\,y=\displaystyle\int\sec^2x\,\mathrm dx=\tan x+C](https://tex.z-dn.net/?f=%5Cimplies%5Csec%20x%5C%2Cy%3D%5Cdisplaystyle%5Cint%5Csec%5E2x%5C%2C%5Cmathrm%20dx%3D%5Ctan%20x%2BC)
![\implies y=\cos x\tan x+C\cos x](https://tex.z-dn.net/?f=%5Cimplies%20y%3D%5Ccos%20x%5Ctan%20x%2BC%5Ccos%20x)
![y=\sin x+C\cos x](https://tex.z-dn.net/?f=y%3D%5Csin%20x%2BC%5Ccos%20x)
- - -
In case the steps where we multiply or divide through by a certain factor weren't clear enough, those steps follow from the procedure for finding an integrating factor. We start with the linear equation
![a(x)y'(x)+b(x)y(x)=c(x)](https://tex.z-dn.net/?f=a%28x%29y%27%28x%29%2Bb%28x%29y%28x%29%3Dc%28x%29)
then rewrite it as
![y'(x)=\dfrac{b(x)}{a(x)}y(x)=\dfrac{c(x)}{a(x)}\iff y'(x)+P(x)y(x)=Q(x)](https://tex.z-dn.net/?f=y%27%28x%29%3D%5Cdfrac%7Bb%28x%29%7D%7Ba%28x%29%7Dy%28x%29%3D%5Cdfrac%7Bc%28x%29%7D%7Ba%28x%29%7D%5Ciff%20y%27%28x%29%2BP%28x%29y%28x%29%3DQ%28x%29)
The integrating factor is a function
![\mu(x)](https://tex.z-dn.net/?f=%5Cmu%28x%29)
such that
![\mu(x)y'(x)+\mu(x)P(x)y(x)=(\mu(x)y(x))'](https://tex.z-dn.net/?f=%5Cmu%28x%29y%27%28x%29%2B%5Cmu%28x%29P%28x%29y%28x%29%3D%28%5Cmu%28x%29y%28x%29%29%27)
which requires that
![\mu(x)P(x)=\mu'(x)](https://tex.z-dn.net/?f=%5Cmu%28x%29P%28x%29%3D%5Cmu%27%28x%29)
This is a separable ODE, so solving for
![\mu](https://tex.z-dn.net/?f=%5Cmu)
we have
![\mu(x)P(x)=\dfrac{\mathrm d\mu(x)}{\mathrm dx}\iff\dfrac{\mathrm d\mu(x)}{\mu(x)}=P(x)\,\mathrm dx](https://tex.z-dn.net/?f=%5Cmu%28x%29P%28x%29%3D%5Cdfrac%7B%5Cmathrm%20d%5Cmu%28x%29%7D%7B%5Cmathrm%20dx%7D%5Ciff%5Cdfrac%7B%5Cmathrm%20d%5Cmu%28x%29%7D%7B%5Cmu%28x%29%7D%3DP%28x%29%5C%2C%5Cmathrm%20dx)
![\implies\ln|\mu(x)|=\displaystyle\int P(x)\,\mathrm dx](https://tex.z-dn.net/?f=%5Cimplies%5Cln%7C%5Cmu%28x%29%7C%3D%5Cdisplaystyle%5Cint%20P%28x%29%5C%2C%5Cmathrm%20dx)
![\implies\mu(x)=\exp\left(\displaystyle\int P(x)\,\mathrm dx\right)](https://tex.z-dn.net/?f=%5Cimplies%5Cmu%28x%29%3D%5Cexp%5Cleft%28%5Cdisplaystyle%5Cint%20P%28x%29%5C%2C%5Cmathrm%20dx%5Cright%29)
and so on.
Formula for area of trapezoid is (base a + base b)*h/2
A = (a+b)*h/2
1833 = (52+42)*h/2
2*1833=94h
3666=94h
h = 3666/94 = 39
the height of trapezoid is equal 39 meters
hope helped
12/99 is the fraction form