Answer:
by my degree in math and other things this is correct.
Answer:
The option is first one that is
![\frac{m^{7}n^{3}n}{m}](https://tex.z-dn.net/?f=%5Cfrac%7Bm%5E%7B7%7Dn%5E%7B3%7Dn%7D%7Bm%7D)
Step-by-step explanation:
Given:
![\frac{m^{7}n^{3}}{mn^{-1} },m\neq 0,n\neq 0](https://tex.z-dn.net/?f=%5Cfrac%7Bm%5E%7B7%7Dn%5E%7B3%7D%7D%7Bmn%5E%7B-1%7D%20%7D%2Cm%5Cneq%200%2Cn%5Cneq%200)
After negative exponent eliminated we get
![\frac{m^{7}n^{3}n}{m}](https://tex.z-dn.net/?f=%5Cfrac%7Bm%5E%7B7%7Dn%5E%7B3%7Dn%7D%7Bm%7D)
Negative exponent :
The variable containing negative powers. Here the variable( n⁻¹) is negative exponent.
Law of indices
![a^{-1} = \frac{1}{a}\\\\Here\\n^{-1} = \frac{1}{n}\\\\](https://tex.z-dn.net/?f=a%5E%7B-1%7D%20%3D%20%5Cfrac%7B1%7D%7Ba%7D%5C%5C%5C%5CHere%5C%5Cn%5E%7B-1%7D%20%3D%20%5Cfrac%7B1%7D%7Bn%7D%5C%5C%5C%5C)
![\\\textrm{Using law of indices we get}\\\frac{m^{7}n^{3}}{mn^{-1} }=\frac{m^{7}n^{3}}{m\frac{1}{n} } }\\\\ \frac{m^{7}n^{3}}{mn^{-1} }=\frac{m^{7}n^{3}n}{m}](https://tex.z-dn.net/?f=%5C%5C%5Ctextrm%7BUsing%20law%20of%20indices%20we%20get%7D%5C%5C%5Cfrac%7Bm%5E%7B7%7Dn%5E%7B3%7D%7D%7Bmn%5E%7B-1%7D%20%7D%3D%5Cfrac%7Bm%5E%7B7%7Dn%5E%7B3%7D%7D%7Bm%5Cfrac%7B1%7D%7Bn%7D%20%7D%20%7D%5C%5C%5C%5C%20%5Cfrac%7Bm%5E%7B7%7Dn%5E%7B3%7D%7D%7Bmn%5E%7B-1%7D%20%7D%3D%5Cfrac%7Bm%5E%7B7%7Dn%5E%7B3%7Dn%7D%7Bm%7D)
a. Substitute the given solutions and their derivatives into the ODE.
![y_1=x\implies {y_1}'=1\implies{y_1}''=0](https://tex.z-dn.net/?f=y_1%3Dx%5Cimplies%20%7By_1%7D%27%3D1%5Cimplies%7By_1%7D%27%27%3D0)
![x^2y''-xy'+y=-x+x=0](https://tex.z-dn.net/?f=x%5E2y%27%27-xy%27%2By%3D-x%2Bx%3D0)
![y_2=x\ln x\implies{y_1}'=\ln x+1\implies{y_1}''=\dfrac1x](https://tex.z-dn.net/?f=y_2%3Dx%5Cln%20x%5Cimplies%7By_1%7D%27%3D%5Cln%20x%2B1%5Cimplies%7By_1%7D%27%27%3D%5Cdfrac1x)
![x^2y''-xy'+y=x-x(\ln x+1)+x\ln x=0](https://tex.z-dn.net/?f=x%5E2y%27%27-xy%27%2By%3Dx-x%28%5Cln%20x%2B1%29%2Bx%5Cln%20x%3D0)
Both solutions satisfy the ODE.
b. The Wronskian determinant is
![\begin{vmatrix}x&x\ln x\\1&\ln x+1\end{vmatrix}=x(\ln x+1)-x\ln x=x\neq0](https://tex.z-dn.net/?f=%5Cbegin%7Bvmatrix%7Dx%26x%5Cln%20x%5C%5C1%26%5Cln%20x%2B1%5Cend%7Bvmatrix%7D%3Dx%28%5Cln%20x%2B1%29-x%5Cln%20x%3Dx%5Cneq0)
so the solutions are indeed independent.
c. The ODE has general solution
. Then with the given initial conditions, the constants satisfy
![y(1)=7\implies 7=C_1](https://tex.z-dn.net/?f=y%281%29%3D7%5Cimplies%207%3DC_1)
![y'(1)=2\implies2=C_1+C_2\implies C_2=-5](https://tex.z-dn.net/?f=y%27%281%29%3D2%5Cimplies2%3DC_1%2BC_2%5Cimplies%20C_2%3D-5)
So the ODE has the particular solution,
![\boxed{y(t)=7x-5x\ln x}](https://tex.z-dn.net/?f=%5Cboxed%7By%28t%29%3D7x-5x%5Cln%20x%7D)
Step-by-step explanation:
![\frac{2}{3} \: in \: 10hours](https://tex.z-dn.net/?f=%20%5Cfrac%7B2%7D%7B3%7D%20%20%5C%3A%20in%20%5C%3A%2010hours)
![in \: 1 \: hour \: = \frac{2}{3} \times \frac{1}{10 \: hours}](https://tex.z-dn.net/?f=in%20%5C%3A%201%20%5C%3A%20hour%20%5C%3A%20%20%3D%20%20%5Cfrac%7B2%7D%7B3%7D%20%20%5Ctimes%20%20%5Cfrac%7B1%7D%7B10%20%5C%3A%20hours%7D%20)
![\: \: \: \: \: \: \: = \frac{1}{3} \times \frac{1}{5}](https://tex.z-dn.net/?f=%20%5C%3A%20%20%5C%3A%20%20%5C%3A%20%20%5C%3A%20%20%5C%3A%20%20%5C%3A%20%20%5C%3A%20%20%3D%20%20%5Cfrac%7B1%7D%7B3%7D%20%20%5Ctimes%20%20%5Cfrac%7B1%7D%7B5%7D%20)
![\: \: \: \: \: \: \: = \frac{1}{15}](https://tex.z-dn.net/?f=%20%5C%3A%20%20%5C%3A%20%20%5C%3A%20%20%5C%3A%20%20%5C%3A%20%20%5C%3A%20%20%5C%3A%20%20%3D%20%20%5Cfrac%7B1%7D%7B15%7D%20)
In a day we have 24 hours
So,
![\: \: \: \: \: \: \: \: = 24 \: \times \frac{1}{15}](https://tex.z-dn.net/?f=%20%5C%3A%20%20%5C%3A%20%20%5C%3A%20%20%5C%3A%20%20%5C%3A%20%20%5C%3A%20%20%5C%3A%20%20%5C%3A%20%20%3D%2024%20%20%5C%3A%20%20%5Ctimes%20%20%5Cfrac%7B1%7D%7B15%7D%20)
![\: \: \: \: \: \: \: \: = \frac{8}{5} \: \: gallon \: of \: water](https://tex.z-dn.net/?f=%20%5C%3A%20%20%5C%3A%20%20%5C%3A%20%20%5C%3A%20%20%5C%3A%20%20%5C%3A%20%20%5C%3A%20%20%5C%3A%20%20%3D%20%20%5Cfrac%7B8%7D%7B5%7D%20%20%5C%3A%20%5C%3A%20gallon%20%5C%3A%20%20of%20%5C%3A%20water)
Hope this helps u.... ^_^❤️
No, they are not. !0/20 would be equivalent to 1/2