Answer with explanation:
--------------------------------------------------------Dividing both sides by 8 x
This Integration is of the form ⇒y'+p y=q,which is Linear differential equation.
Integrating Factor
Multiplying both sides by Integrating Factor
![x^{\frac{1}{8}}\times e^{\frac{x^2}{2}}\times [y'+y\times\frac{1+4x^2}{8x}]=\frac{1}{8}\times x^{\frac{1}{8}}\times e^{\frac{x^2}{2}}\\\\ \text{Integrating both sides}\\\\y\times x^{\frac{1}{8}}\times e^{\frac{x^2}{2}}=\frac{1}{8}\int {x^{\frac{1}{8}}\times e^{\frac{x^2}{2}}} \, dx \\\\8y\times x^{\frac{1}{8}}\times e^{\frac{x^2}{2}}=\int {x^{\frac{1}{8}}\times e^{\frac{x^2}{2}}} \, dx\\\\8y\times x^{\frac{1}{8}}\times e^{\frac{x^2}{2}}=-[x^{\frac{9}{8}}]\times\frac{ \Gamma(0.5625, -x^2)}{(-x^2)^{\frac{9}{16}}}\\\\8y\times x^{\frac{1}{8}}\times e^{\frac{x^2}{2}}=(-1)^{\frac{-1}{8}}[ \Gamma(0.5625, -x^2)]+C-----(1)](https://tex.z-dn.net/?f=x%5E%7B%5Cfrac%7B1%7D%7B8%7D%7D%5Ctimes%20e%5E%7B%5Cfrac%7Bx%5E2%7D%7B2%7D%7D%5Ctimes%20%5By%27%2By%5Ctimes%5Cfrac%7B1%2B4x%5E2%7D%7B8x%7D%5D%3D%5Cfrac%7B1%7D%7B8%7D%5Ctimes%20x%5E%7B%5Cfrac%7B1%7D%7B8%7D%7D%5Ctimes%20e%5E%7B%5Cfrac%7Bx%5E2%7D%7B2%7D%7D%5C%5C%5C%5C%20%5Ctext%7BIntegrating%20both%20sides%7D%5C%5C%5C%5Cy%5Ctimes%20x%5E%7B%5Cfrac%7B1%7D%7B8%7D%7D%5Ctimes%20e%5E%7B%5Cfrac%7Bx%5E2%7D%7B2%7D%7D%3D%5Cfrac%7B1%7D%7B8%7D%5Cint%20%7Bx%5E%7B%5Cfrac%7B1%7D%7B8%7D%7D%5Ctimes%20e%5E%7B%5Cfrac%7Bx%5E2%7D%7B2%7D%7D%7D%20%5C%2C%20dx%20%5C%5C%5C%5C8y%5Ctimes%20x%5E%7B%5Cfrac%7B1%7D%7B8%7D%7D%5Ctimes%20e%5E%7B%5Cfrac%7Bx%5E2%7D%7B2%7D%7D%3D%5Cint%20%7Bx%5E%7B%5Cfrac%7B1%7D%7B8%7D%7D%5Ctimes%20e%5E%7B%5Cfrac%7Bx%5E2%7D%7B2%7D%7D%7D%20%5C%2C%20dx%5C%5C%5C%5C8y%5Ctimes%20x%5E%7B%5Cfrac%7B1%7D%7B8%7D%7D%5Ctimes%20e%5E%7B%5Cfrac%7Bx%5E2%7D%7B2%7D%7D%3D-%5Bx%5E%7B%5Cfrac%7B9%7D%7B8%7D%7D%5D%5Ctimes%5Cfrac%7B%20%5CGamma%280.5625%2C%20-x%5E2%29%7D%7B%28-x%5E2%29%5E%7B%5Cfrac%7B9%7D%7B16%7D%7D%7D%5C%5C%5C%5C8y%5Ctimes%20x%5E%7B%5Cfrac%7B1%7D%7B8%7D%7D%5Ctimes%20e%5E%7B%5Cfrac%7Bx%5E2%7D%7B2%7D%7D%3D%28-1%29%5E%7B%5Cfrac%7B-1%7D%7B8%7D%7D%5B%20%5CGamma%280.5625%2C%20-x%5E2%29%5D%2BC-----%281%29)
When , x=1, gives , y=9.
Evaluate the value of C and substitute in the equation 1.
Answer- your answer would be 5/11
To solve this problem, we need to first find the dimensions of the side of the blue and purple squares.
We're given that the purple (smaller) square has a side length of x inches.
We are also given that the blue band has a width of 5 inches.
Since the blue band surrounds the purple square on both sides, the length of the blue square is x+2(5)=x+10 inches.
The net area of the band is therefore the difference of the area of the blue square and the purple square, namely take out the area of the purple square from the blue.
Therefore
Area of band
[recall
or 20(x+5) if you wish.
