Answer: The required solution of the given differential equation is
Step-by-step explanation: We are given to solve the following differential equation :
Let, 
be an auxiliary solution of equation (i).
Then, 
Substituting these values in equation (i), we get
![m^3e^{mx}+4m^2e^{mx}-16me^{mx}-64e^{mx}=0\\\\\Rightarrow (m^3+4m^2-16m-64)e^{mx}=0\\\\\Rightarrow m^3+4m^2-16m-64=0,~~~~~~~~~[\textup{since }e^{mx}\neq 0]\\\\\Rightarrow m^2(m-4)+8m(m-4)+16(m-4)=0\\\\\Rightarrow (m-4)(m^2+8m+16)=0\\\\\Rightarrow (m-4)(m+4)^2=0\\\\\Rightarrow m-4=0,~~(m+4)^2=0\\\\\Rightarrow m=4,~m=-4,~-4.](https://tex.z-dn.net/?f=m%5E3e%5E%7Bmx%7D%2B4m%5E2e%5E%7Bmx%7D-16me%5E%7Bmx%7D-64e%5E%7Bmx%7D%3D0%5C%5C%5C%5C%5CRightarrow%20%28m%5E3%2B4m%5E2-16m-64%29e%5E%7Bmx%7D%3D0%5C%5C%5C%5C%5CRightarrow%20m%5E3%2B4m%5E2-16m-64%3D0%2C~~~~~~~~~%5B%5Ctextup%7Bsince%20%7De%5E%7Bmx%7D%5Cneq%200%5D%5C%5C%5C%5C%5CRightarrow%20m%5E2%28m-4%29%2B8m%28m-4%29%2B16%28m-4%29%3D0%5C%5C%5C%5C%5CRightarrow%20%28m-4%29%28m%5E2%2B8m%2B16%29%3D0%5C%5C%5C%5C%5CRightarrow%20%28m-4%29%28m%2B4%29%5E2%3D0%5C%5C%5C%5C%5CRightarrow%20m-4%3D0%2C~~%28m%2B4%29%5E2%3D0%5C%5C%5C%5C%5CRightarrow%20m%3D4%2C~m%3D-4%2C~-4.)
So, the general solution is given by
Then, we have
With the conditions given, we get
![y^\prime(0)=4A-4B+C\\\\\Rightarrow 4A-4B+C=26\\\\\Rightarrow 4(A+A)+C=26~~~~~~~~~~~~~~~~[\textup{using equation (i)}]\\\\\Rightarrow C=26-8A~~~~~~~~~~~~~~~~~~~~~~~~~~~(iii)](https://tex.z-dn.net/?f=y%5E%5Cprime%280%29%3D4A-4B%2BC%5C%5C%5C%5C%5CRightarrow%204A-4B%2BC%3D26%5C%5C%5C%5C%5CRightarrow%204%28A%2BA%29%2BC%3D26~~~~~~~~~~~~~~~~%5B%5Ctextup%7Busing%20equation%20%28i%29%7D%5D%5C%5C%5C%5C%5CRightarrow%20C%3D26-8A~~~~~~~~~~~~~~~~~~~~~~~~~~~%28iii%29)
and
![y^{\prime\prime}(0)=16A+16B-8C\\\\\Rightarrow 16A-16A-8C=-16~~~~~~~~~~~~[\textup{using equation (ii)}]\\\\\Rightarrow -8C=-16\\\\\Rightarrow C=2.](https://tex.z-dn.net/?f=y%5E%7B%5Cprime%5Cprime%7D%280%29%3D16A%2B16B-8C%5C%5C%5C%5C%5CRightarrow%2016A-16A-8C%3D-16~~~~~~~~~~~~%5B%5Ctextup%7Busing%20equation%20%28ii%29%7D%5D%5C%5C%5C%5C%5CRightarrow%20-8C%3D-16%5C%5C%5C%5C%5CRightarrow%20C%3D2.)
From equation (iii), we get
From equation (ii), we get
Therefore, the required solution of the given differential equation is
Answer:
Cant do this because we dont have your username and password and the link you gave us isnt signed into your account sorry
The height of the isosceles triangle is 8.49 inches.
<h3>
How to find the height of the triangle?</h3>
Here we have a triangle such that two of the sides measure 9 inches, and the base measures 6 inches.
So this is an isosceles triangle.
We can divide the isosceles triangle into two smaller right triangles, such that the side that measures 9 inches is the hypotenuse, the base is 3 inches, and the height of the isosceles triangle is the other cathetus.
By Pythagorean's theorem, we can write:
(9in)^2 = (3 in)^2 + h^2
Where h is the height that we are trying to find.
Solving that for h we get:
h = √( (9 in)^2 - (3in)^2) = 8.49 inches.
We conclude that the height of the isosceles triangle is 8.49 inches.
If you want to learn more about triangles:
brainly.com/question/2217700
#SPJ1
The answer will be 46,200 Hopes this helps. :)
