Solving the Differential Equation: y' = y^4 - 6y^3 + 5y^2

[semc]

A function y(t) satises the differential equation

$$\frac{dy}{dx}$$ = y⁴ - 6y³ + 5y²

What are the constant solutions of the equation?I have totally no idea what the question is asking can someone explain why substituting y with a constant solves it?

 

[leto78]

When you are differentiating y(t) in terms of x, the variable t is not employed. Thus, all the terms y are constant.

dy(t)/dx=y⁴ - 6y³ + 5y²
is solved as
y(t)=(y⁴ - 6y³ + 5y²)x+k

where k is constant.

On the other hand, if the question was
dy/dt=y⁴ - 6y³ + 5y²

This would be a 5th degree differential equation... maybe solvable.

 

[payumooli]

y = constant
$$\frac{dy}{dx} = 0$$
y⁴ - 6y³ + 5y² = 0

 

[payumooli]

solve the polynomial in the above post to get the constant solutions

 

[semc]

Actually I don't really understand the question. I was hoping someone can explain to me what it actually means. If they want to make y a constant then why did they differentiate y with respect to x and get this equation?

 

[leto78]

I think payumooli interpretation is probably the most accurate. It is true that the question is poorly formed.
If by constants they mean the zeros, than it is simple.
If you assume that
$$\ \frac{dy(t)}{dx} = 0$$
then the equation is
$$y^4 - 6y^3 + 5y^2=0$$

which is solved by $$y^2(5/6 - y)=0$$ where $$y=0$$ or $$y=5/6$$


If you assume that there was a two-variable function y(x,t), where
$$\ \frac{dy(x,t)}{dx} = y^4 - 6y^3 + 5y^2$$
the case is a bit different. Here, the constants of y(x,t) in relation to x is the right hand side of the equation, since:
$$\ \frac{d(kx+c)}{dx}=k$$