Answer:
b
Step-by-step explanation:
cant explain
The answer would be x equals 4
Check if the equation is exact, which happens for ODEs of the form
![M(x,y)\,\mathrm dx+N(x,y)\,\mathrm dy=0](https://tex.z-dn.net/?f=M%28x%2Cy%29%5C%2C%5Cmathrm%20dx%2BN%28x%2Cy%29%5C%2C%5Cmathrm%20dy%3D0)
if
.
We have
![M(x,y)=x^2+y^2+x\implies\dfrac{\partial M}{\partial y}=2y](https://tex.z-dn.net/?f=M%28x%2Cy%29%3Dx%5E2%2By%5E2%2Bx%5Cimplies%5Cdfrac%7B%5Cpartial%20M%7D%7B%5Cpartial%20y%7D%3D2y)
![N(x,y)=xy\implies\dfrac{\partial N}{\partial x}=y](https://tex.z-dn.net/?f=N%28x%2Cy%29%3Dxy%5Cimplies%5Cdfrac%7B%5Cpartial%20N%7D%7B%5Cpartial%20x%7D%3Dy)
so the ODE is not quite exact, but we can find an integrating factor
so that
![\mu(x,y)M(x,y)\,\mathrm dx+\mu(x,y)N(x,y)\,\mathrm dy=0](https://tex.z-dn.net/?f=%5Cmu%28x%2Cy%29M%28x%2Cy%29%5C%2C%5Cmathrm%20dx%2B%5Cmu%28x%2Cy%29N%28x%2Cy%29%5C%2C%5Cmathrm%20dy%3D0)
<em>is</em> exact, which would require
![\dfrac{\partial(\mu M)}{\partial y}=\dfrac{\partial(\mu N)}{\partial x}\implies \dfrac{\partial\mu}{\partial y}M+\mu\dfrac{\partial M}{\partial y}=\dfrac{\partial\mu}{\partial x}N+\mu\dfrac{\partial N}{\partial x}](https://tex.z-dn.net/?f=%5Cdfrac%7B%5Cpartial%28%5Cmu%20M%29%7D%7B%5Cpartial%20y%7D%3D%5Cdfrac%7B%5Cpartial%28%5Cmu%20N%29%7D%7B%5Cpartial%20x%7D%5Cimplies%20%5Cdfrac%7B%5Cpartial%5Cmu%7D%7B%5Cpartial%20y%7DM%2B%5Cmu%5Cdfrac%7B%5Cpartial%20M%7D%7B%5Cpartial%20y%7D%3D%5Cdfrac%7B%5Cpartial%5Cmu%7D%7B%5Cpartial%20x%7DN%2B%5Cmu%5Cdfrac%7B%5Cpartial%20N%7D%7B%5Cpartial%20x%7D)
![\implies\mu\left(\dfrac{\partial N}{\partial x}-\dfrac{\partial M}{\partial y}\right)=M\dfrac{\partial\mu}{\partial y}-N\dfrac{\partial\mu}{\partial x}](https://tex.z-dn.net/?f=%5Cimplies%5Cmu%5Cleft%28%5Cdfrac%7B%5Cpartial%20N%7D%7B%5Cpartial%20x%7D-%5Cdfrac%7B%5Cpartial%20M%7D%7B%5Cpartial%20y%7D%5Cright%29%3DM%5Cdfrac%7B%5Cpartial%5Cmu%7D%7B%5Cpartial%20y%7D-N%5Cdfrac%7B%5Cpartial%5Cmu%7D%7B%5Cpartial%20x%7D)
Notice that
![\dfrac{\partial N}{\partial x}-\dfrac{\partial M}{\partial y}=y-2y=-y](https://tex.z-dn.net/?f=%5Cdfrac%7B%5Cpartial%20N%7D%7B%5Cpartial%20x%7D-%5Cdfrac%7B%5Cpartial%20M%7D%7B%5Cpartial%20y%7D%3Dy-2y%3D-y)
is independent of <em>x</em>, and dividing this by
gives an expression independent of <em>y</em>. If we assume
is a function of <em>x</em> alone, then
, and the partial differential equation above gives
![-\mu y=-xy\dfrac{\mathrm d\mu}{\mathrm dx}](https://tex.z-dn.net/?f=-%5Cmu%20y%3D-xy%5Cdfrac%7B%5Cmathrm%20d%5Cmu%7D%7B%5Cmathrm%20dx%7D)
which is separable and we can solve for
easily.
![-\mu=-x\dfrac{\mathrm d\mu}{\mathrm dx}](https://tex.z-dn.net/?f=-%5Cmu%3D-x%5Cdfrac%7B%5Cmathrm%20d%5Cmu%7D%7B%5Cmathrm%20dx%7D)
![\dfrac{\mathrm d\mu}\mu=\dfrac{\mathrm dx}x](https://tex.z-dn.net/?f=%5Cdfrac%7B%5Cmathrm%20d%5Cmu%7D%5Cmu%3D%5Cdfrac%7B%5Cmathrm%20dx%7Dx)
![\ln|\mu|=\ln|x|](https://tex.z-dn.net/?f=%5Cln%7C%5Cmu%7C%3D%5Cln%7Cx%7C)
![\implies \mu=x](https://tex.z-dn.net/?f=%5Cimplies%20%5Cmu%3Dx)
So, multiply the original ODE by <em>x</em> on both sides:
![(x^3+xy^2+x^2)\,\mathrm dx+x^2y\,\mathrm dy=0](https://tex.z-dn.net/?f=%28x%5E3%2Bxy%5E2%2Bx%5E2%29%5C%2C%5Cmathrm%20dx%2Bx%5E2y%5C%2C%5Cmathrm%20dy%3D0)
Now
![\dfrac{\partial(x^3+xy^2+x^2)}{\partial y}=2xy](https://tex.z-dn.net/?f=%5Cdfrac%7B%5Cpartial%28x%5E3%2Bxy%5E2%2Bx%5E2%29%7D%7B%5Cpartial%20y%7D%3D2xy)
![\dfrac{\partial(x^2y)}{\partial x}=2xy](https://tex.z-dn.net/?f=%5Cdfrac%7B%5Cpartial%28x%5E2y%29%7D%7B%5Cpartial%20x%7D%3D2xy)
so the modified ODE is exact.
Now we look for a solution of the form
, with differential
![\mathrm dF=\dfrac{\partial F}{\partial x}\,\mathrm dx+\dfrac{\partial F}{\partial y}\,\mathrm dy=0](https://tex.z-dn.net/?f=%5Cmathrm%20dF%3D%5Cdfrac%7B%5Cpartial%20F%7D%7B%5Cpartial%20x%7D%5C%2C%5Cmathrm%20dx%2B%5Cdfrac%7B%5Cpartial%20F%7D%7B%5Cpartial%20y%7D%5C%2C%5Cmathrm%20dy%3D0)
The solution <em>F</em> satisfies
![\dfrac{\partial F}{\partial x}=x^3+xy^2+x^2](https://tex.z-dn.net/?f=%5Cdfrac%7B%5Cpartial%20F%7D%7B%5Cpartial%20x%7D%3Dx%5E3%2Bxy%5E2%2Bx%5E2)
![\dfrac{\partial F}{\partial y}=x^2y](https://tex.z-dn.net/?f=%5Cdfrac%7B%5Cpartial%20F%7D%7B%5Cpartial%20y%7D%3Dx%5E2y)
Integrating both sides of the first equation with respect to <em>x</em> gives
![F(x,y)=\dfrac{x^4}4+\dfrac{x^2y^2}2+\dfrac{x^3}3+f(y)](https://tex.z-dn.net/?f=F%28x%2Cy%29%3D%5Cdfrac%7Bx%5E4%7D4%2B%5Cdfrac%7Bx%5E2y%5E2%7D2%2B%5Cdfrac%7Bx%5E3%7D3%2Bf%28y%29)
Differentiating both sides with respect to <em>y</em> gives
![\dfrac{\partial F}{\partial y}=x^2y+\dfrac{\mathrm df}{\mathrm dy}=x^2y](https://tex.z-dn.net/?f=%5Cdfrac%7B%5Cpartial%20F%7D%7B%5Cpartial%20y%7D%3Dx%5E2y%2B%5Cdfrac%7B%5Cmathrm%20df%7D%7B%5Cmathrm%20dy%7D%3Dx%5E2y)
![\implies\dfrac{\mathrm df}{\mathrm dy}=0\implies f(y)=C](https://tex.z-dn.net/?f=%5Cimplies%5Cdfrac%7B%5Cmathrm%20df%7D%7B%5Cmathrm%20dy%7D%3D0%5Cimplies%20f%28y%29%3DC)
So the solution to the ODE is
![F(x,y)=C\iff \dfrac{x^4}4+\dfrac{x^2y^2}2+\dfrac{x^3}3+C=C](https://tex.z-dn.net/?f=F%28x%2Cy%29%3DC%5Ciff%20%5Cdfrac%7Bx%5E4%7D4%2B%5Cdfrac%7Bx%5E2y%5E2%7D2%2B%5Cdfrac%7Bx%5E3%7D3%2BC%3DC)
![\implies\boxed{\dfrac{x^4}4+\dfrac{x^2y^2}2+\dfrac{x^3}3=C}](https://tex.z-dn.net/?f=%5Cimplies%5Cboxed%7B%5Cdfrac%7Bx%5E4%7D4%2B%5Cdfrac%7Bx%5E2y%5E2%7D2%2B%5Cdfrac%7Bx%5E3%7D3%3DC%7D)
People moved to the cities to escape war. Some people had left the countryside to work in the arms industry and some had left for Rome looking for subsistence. Rome became the most populous city in Europe and West Asia.
let "a number" = x
(x) + (x + 1) + (x + 2) = 72
Simplify. Combine like terms
3x + 3 = 72
Isolate the x. Note the equal sign. What you do to one side, you do to the other. Do the opposite of PEMDAS.
3x + 3 (-3) = 72 (-3)
3x = 72 - 3
3x = 69
3x/3 = 69/3
x = 69/3
x = 23
23 is the smallest number
hope this helps