A little help (should be simple)

[tumbleweed]

I have a homework problem I am stuck on. I cannot find examples in my book that are similar. It should be fairly simple since I am not too far into the course. Here it is:

Solve dy/dt = y^2 - 2yt + t^2 + y - t + 1. Find all solutions that satisfy y(0) = -1

I know it simplifies to dy/dt = (y-t)^2 + (y-t) + 1, but I don't know if I'm on the right track with that or not.

Any help? I don't really know how to do non-linear problems....

 

[galactus]

Actually, this problem may not be that easy for a beginning problem.

Maybe try a substitution for y-t. Say, u=y-t.

Take a look at \(\displaystyle y=t-1\) as a solution.

 

[Deleted member 4993]

I have a homework problem I am stuck on. I cannot find examples in my book that are similar. It should be fairly simple since I am not too far into the course. Here it is:

Solve dy/dt = y^2 - 2yt + t^2 + y - t + 1. Find all solutions that satisfy y(0) = -1

I know it simplifies to dy/dt = (y-t)^2 + (y-t) + 1, but I don't know if I'm on the right track with that or not.

Any help? I don't really know how to do non-linear problems....

Following Galactus's hint:

u = y - t

du/dt = dy/dt -1

then

dy/dt = (y-t)^2 + (y-t) + 1

dy/dt - 1= (y-t)^2 + (y-t)

du/dt = u² + u

du/[u(u+1)] = dt

du/u - du/(u+1) = dt

Integrating both sides, we get:

ln[u/(u+1)] = t + C

Now finish it....

 

[burakaltr]

I have a homework problem I am stuck on. I cannot find examples in my book that are similar. It should be fairly simple since I am not too far into the course. Here it is:

Solve dy/dt = y^2 - 2yt + t^2 + y - t + 1. Find all solutions that satisfy y(0) = -1

I know it simplifies to dy/dt = (y-t)^2 + (y-t) + 1, but I don't know if I'm on the right track with that or not.

Any help? I don't really know how to do non-linear problems....

dy/dt=(y-t)**2 + ( y-t ) + 1

y=x+t
dy=dx+dt

dx/dt + 1 = x**2 + x + 1

dx/dt = x**2 + x

dx/[(x)(x+1)]=dt

(1/x ) (1/[x+1])=[A/x] + [B /(x+1)]

1=Ax+A+Bx

A=1

Ax+Bx=0 x

A=-B

B=-1

ln(abs(x)) - ln(abs(x+1)) = t

ln (abs(x / (x+1)) =t

e**t= x/(x+1) for e**t must be pos.

x=y-t

e**t=(y-t)/(y-t+1)

e**2 y - t e**t + e**t = y-t

( e**2 - 1)y=(e**t)(-t-1) - t

y=-([(e**t)(t+1)] + t) /[e**2 - 1]


y=[(e**t)(t+1)+t] / (1- e**2)