Answer with explanation:
The given differential equation is
y" -y'+y=2 sin 3x------(1)
Let, y'=z
y"=z'

Substituting the value of , y, y' and y" in equation (1)
z'-z+zx=2 sin 3 x
z'+z(x-1)=2 sin 3 x-----------(1)
This is a type of linear differential equation.
Integrating factor

Multiplying both sides of equation (1) by integrating factor and integrating we get


Answer:
100,000
Step-by-step explanation:
I'm pretty sure. hope this helped <3 also I LOVE UR PFP ;)
Answer:
b and d
Step-by-step explanation:
they both have b
Do you want it to multiply. O que ?
Answer: it is d becaues that is the mode