Solving a Non-Linear Differential Equation Using Exact Equations

[Tom1]

Hi,

I am trying to:
Show that y1(t)=t^2 and y2(t)=t^2+1

for the differential equation:
dy/dt=-y^2+y+2t^2y+2t-t^2-t^4

This is clearly a non-linear, non-seperable equation, and I cannot see how to try the method of exact equations on it. Therefore, I am stuck.

Anyone have any ideas of what to try?

 

[sutupidmath]

are you trying to show that y1(t)=t^2 and y2(t)=t^2+1
are two solutions of that diffl. eq or, just to solve that diff. eq? if the former is the case, all you need to do is find the derivative of y1, y2 and plug in the diff eq and see if the right hand is equal to the left hand. if the latter than ther is more work to do!

 

[Tom1]

Trying to show that those are solutions to the diff. eq.

If I could solve the diff. eq. I guess it would show that those are particular solutions though.

 

[sutupidmath]

Show that y1(t)=t^2 and y2(t)=t^2+1

for the differential equation:
dy/dt=-y^2+y+2t^2y+2t-t^2-t^4
y1'=2t, so

2t=-(t^2)^2+t^2+2t^2*t^2+2t-t^2-t^4=-t^4+2t^4 +2t-t^4=2t, so this is a solution. try the same thing with y2

 

[Tom1]

Thank you much.

 

[Rainbow Child]

This equation is a Riccati type equation, i.e. $y'(t)=a(t)\,y(t)^2+b(t)\,y(t)+c(t)$. The general solution can be found if one knows a particular solution $y_p(t)$ by the transformation $y(t)=y_p(t)+\frac{1}{u(t)}$ which makes it a linear one.
For the problem at hand if you write $y(t)=t^2+\frac{1}{u(t)}$ then the original ODE reduces to $u'(t)=1-u(t)\Rightarrow u(t)=C\,e^{-t}+1$.