For the following rectangular prism, drag and drop the steps needed to find the volume in the correct order. Then type the volum
e in the last box. Not all pieces will be used.
Picture shown.
Picture shown.
1 answer:
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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