Help me - A function satisfies the differential equation

Join the discussion
Registration is free. Ask a follow-up in this thread, or start your own.
5 replies · 7K views
butbi9x
Messages
1
Reaction score
0
A function satisfies the differential equation:

[tex]\frac{dy}{dt}= y^{4}-6y^{3}+5y^{2}[/tex]

a. What are the constant solutions of the equation?
b. For what values of y is y increasing?
c. For what values of y is y decreasing?
 
Physics news on Phys.org
dy/dt=y'(t), its ur function or, simple, y(t), after integration.
 
Last edited:
butbi9x said:
A function satisfies the differential equation:

[tex]\frac{dy}{dt}= y^{4}-6y^{3}+5y^{2}[/tex]

a. What are the constant solutions of the equation?
b. For what values of y is y increasing?
c. For what values of y is y decreasing?
Do you understand that this does NOT require that you actually solve the differential equation? It only requires that you solve an algebraic equation and two inequalities.
 
i think it should be
y'(t)=t^4-6t^3+5t^2
where
y'(t)=dy/dt
now integrate it
we have
y(t)=...
constant solution mean y'(t)=0.
mean no variation w.r.t "t".
for what value of "t" is y increasing...
and for what value of "t" is y decreasing.
i think now u can handle it easily
 
Last edited:
masqau said:
i think it should be
y'(t)=t^4-6t^3+5t^2
No, that's completely wrong. The DE equation is dy/dt = y4 - 6y3 + 5y2. If if were as you have it, this would be a different problem completely.
masqau said:
where
y'(t)=dy/dt
now integrate it
we have
y(t)=...
constant solution mean y'(t)=0.
mean no variation w.r.t "t".
for what value of "t" is y increasing...
and for what value of "t" is y decreasing.
i think now u can handle it easily