# Thread: Right or wrong -- Help me !!!

1. ## Right or wrong -- Help me !!!

* A function y(t) satisfies the differential equation $\displaystyle \frac{dy}{dt}=y^{4}-6y^{3}+5y^{2}$

a. What are the constant solutions of the equation ?
My answer : Solution :$\displaystyle y=\frac{y^{5}}{5}-\frac{6y^{4}}{4}+\frac{5y^{3}}{3}$, is not depend t , therefore y are the constant solutions of the equation.

Right or wrong ?

b. For what values of y is y increasing ? --> Should be "For what values of t is y increasing " Right or worng ?

2. Originally Posted by butbi9x
* A function y(t) satisfies the differential equation $\displaystyle \frac{dy}{dt}=y^{4}-6y^{3}+5y^{2}$

a. What are the constant solutions of the equation ?
My answer : Solution :$\displaystyle y=\frac{y^{5}}{5}-\frac{6y^{4}}{4}+\frac{5y^{3}}{3}$, is not depend t , therefore y are the constant solutions of the equation.

Right or wrong ?

b. For what values of y is y increasing ? --> Should be "For what values of t is y increasing " Right or worng ?
You'll notice that your equation can be written as

$\displaystyle \frac{dy}{dt}=y^2(y-1)(y-5)$

Now the question asks, for what constants (i.e. y = c) is this equation satisfied? Now between these constants either $\displaystyle y' > 0$ increasing or $\displaystyle y' < 0$ decreasing.

3. Originally Posted by butbi9x
* A function y(t) satisfies the differential equation $\displaystyle \frac{dy}{dt}=y^{4}-6y^{3}+5y^{2}$

a. What are the constant solutions of the equation ?
My answer : Solution :$\displaystyle y=\frac{y^{5}}{5}-\frac{6y^{4}}{4}+\frac{5y^{3}}{3}$, is not depend t , therefore y are the constant solutions of the equation.

Right or wrong ?
Sorry, but that makes no sense at all. You cannot simply write an equation with only y and deduce from the equation that y does not depend on t!
Perhaps more importantly, you integrated the left side with respect to t, $\displaystyle \int (\frac{dy}{dt})dt= y$ but integrated the right side with respect to y. You can't do that! If y is a constant then dy/dx= 0. Solve $\displaystyle y^4- 6y^3+ 5y^2= 0$. As danny arrigo pointed out, that can be easily factored.

b. For what values of y is y increasing ? --> Should be "For what values of t is y increasing " Right or worng ?
No, you are not. y will be increasing when dy/dx> 0 and, since you can factor the right hand side, that will happen when an even number of those factors are negative.

$\displaystyle y^2(y-1)(y-5)$ can change sign only at y= 0, y= 1, y= 5. If y< 0, all 4 factors will be negative so the product is positive. If y is between 0 and 1, two factors will be negative so the product is still positive. If y is between 1 and 5, only y- 5 is negative so the product is negative. If y> 5, none of the factors is negative so the product is positive.

4. However, if the equation were

$\displaystyle \frac{dy}{dt}=t^4-6t^3+5t^2$

then we would have

$\displaystyle y=\frac{t^5}{5}-\frac{6t^4}{4}+\frac{5t^3}{3}+C.$

5. True- but it wasn't!