You should use 3 /13 for 4 1/3 and 6/31 for 5 1/6
Hope this helps!
I think it’s 7x plus -4x which equals -3 then u would do 2 plus -7 which equals-5 so it will be positive 8 I think I do t really know I tried
![y'-\dfrac13y=\dfrac13xe^x\ln x\,y^{-2}](https://tex.z-dn.net/?f=y%27-%5Cdfrac13y%3D%5Cdfrac13xe%5Ex%5Cln%20x%5C%2Cy%5E%7B-2%7D)
Divide both sides by
:
![3y^2y'-y^3=xe^x\ln x](https://tex.z-dn.net/?f=3y%5E2y%27-y%5E3%3Dxe%5Ex%5Cln%20x)
Substitute
, so that
.
![v'-v=xe^x\ln x](https://tex.z-dn.net/?f=v%27-v%3Dxe%5Ex%5Cln%20x)
Multiply both sides by
:
![e^{-x}v'-e^{-x}v=x\ln x](https://tex.z-dn.net/?f=e%5E%7B-x%7Dv%27-e%5E%7B-x%7Dv%3Dx%5Cln%20x)
The left side can be condensed into the derivative of a product.
![(e^{-x}v)'=x\ln x](https://tex.z-dn.net/?f=%28e%5E%7B-x%7Dv%29%27%3Dx%5Cln%20x)
Integrate both sides to get
![e^{-x}v=\dfrac12x^2\ln x-\dfrac14x^2+C](https://tex.z-dn.net/?f=e%5E%7B-x%7Dv%3D%5Cdfrac12x%5E2%5Cln%20x-%5Cdfrac14x%5E2%2BC)
Solve for
:
![v=\dfrac12x^2e^x\ln x-\dfrac14x^2e^x+Ce^x](https://tex.z-dn.net/?f=v%3D%5Cdfrac12x%5E2e%5Ex%5Cln%20x-%5Cdfrac14x%5E2e%5Ex%2BCe%5Ex)
Solve for
:
![y^3=\dfrac12x^2e^x\ln x-\dfrac14x^2e^x+Ce^x](https://tex.z-dn.net/?f=y%5E3%3D%5Cdfrac12x%5E2e%5Ex%5Cln%20x-%5Cdfrac14x%5E2e%5Ex%2BCe%5Ex)
![\implies\boxed{y(x)=\sqrt[3]{\dfrac14x^2e^x(2\ln x-1)+Ce^x}}](https://tex.z-dn.net/?f=%5Cimplies%5Cboxed%7By%28x%29%3D%5Csqrt%5B3%5D%7B%5Cdfrac14x%5E2e%5Ex%282%5Cln%20x-1%29%2BCe%5Ex%7D%7D)