Volume of a cilinger, V = π(r^2)H
Here, V = 540π ft^3 and H =15 ft
Then, r^2 = 540π ft^3 / [π15ft] = 36ft^2
r = √(36ft^2) = 6ft
And the diameter is 2*r = 12 ft
Answer:
Therefore the particular solution of the given differential equation is
Step-by-step explanation:
The given ordinary differential equation is
If y₁(x) =x² is a solution of ODE then it will be satisfy the ODE
y₁'(x)= 2x and y₁"(x)=2
Putting the value of y₁'(x) and y₁"(x) in x²y"-3xy'+4y=0 we get
x².2-3x.2x+4x²= 2x²-6x²+4x²=0
Therefore y₁(x) is a solution of the given differential equation.
Again,
y₂(x) =(x²ln x ) is a solution of ODE then it will be satisfy the ODE.
= 3+2 ln x
Putting the value of y₂'(x) and y₂"(x) in x²y"-3xy'+4y=0 we get
=0
Therefore y₂(x) is a solution of the given differential equation.
The wronskian of y₁(x) and y₂(x) is
=x²(2x ln x+x)-x²ln x(2x)
=2x³ ln x +x³ - 2x³ln x
=x³≠0
Here
The particular solution is
Let ln x =u
Therefore the particular solution of the given differential equation is
-1.25 is greater because if u can it like this 1.25 1.03 it will still be greater
Answer:
2x^2-18x
Step-by-step explanation:
Answer:
15x+30
Step-by-step explanation:
multiply 15 and x + multiply 15 and 2
15x+30