Suppose
is another solution. Then

Substituting these derivatives into the ODE gives


Let
, so that

Then the ODE becomes

and we can condense the left hand side as a derivative of a product,
![\dfrac{\mathrm d}{\mathrm dx}[x^5u]=0](https://tex.z-dn.net/?f=%5Cdfrac%7B%5Cmathrm%20d%7D%7B%5Cmathrm%20dx%7D%5Bx%5E5u%5D%3D0)
Integrate both sides with respect to
:
![\displaystyle\int\frac{\mathrm d}{\mathrm dx}[x^5u]\,\mathrm dx=C](https://tex.z-dn.net/?f=%5Cdisplaystyle%5Cint%5Cfrac%7B%5Cmathrm%20d%7D%7B%5Cmathrm%20dx%7D%5Bx%5E5u%5D%5C%2C%5Cmathrm%20dx%3DC)

Solve for
:

Solve for
:

So another linearly independent solution is
.
Area: Trapezoids A = ½h(b1<span> + b</span>2<span>)
</span>Rhombuses A=pq/2
Kites A=pq/2
Slope is already given in the question so we only need to solve for the y-intercept.
Slope intercept form: y = mx + b
m = slope
b = y-intercept
-3/5 = -0.6
1 = -0.6(1) + b
1 = -0.6 + b
1 + 0.6 = -0.6 + b + 0.6
1.6 = b
Now, write in slope-intercept form.
y = -3/5x - 1.6
______
Best of Luck,
Wolfyy :)
The perimeter of the sandbox is 60 ft.
5x5=25
Two sides = 25, so 25+25 = 50.
Two other sides = 5, so 5+5=10
50+10 = 60, and that is the correct answer.
If you would like any more assistance with this type of questions, let me know and I will be able to further assist you. Thank you!