a) The ODE is exact. Using N=x+e^y and M=x^2+y we get
the implicit family of solutions:
(1/3) x^3 + xy + e^y = C.
b) The ODE is linear. In standard form we get p=4/t and g=(1/t^3)e^(-t^2).
Using the integrating factor method we get the explicit solution:
y = [e^(-1) - e^(-t^2)] / [2t^4].
c) After dividing through by sin(x)sin(y) we see that the ODE is separable.
Integrating using "u-substitution" on sin(x) and sin(y) leads to:
sin^2(x) sin(y) = C.
d) Since e^(x+y)=e^(x) e^(y) we see that the ODE is separable.
Integrating leads to the implicit family of solutions:
ye^(-y) + e^(-y) + e^x = 1 + 4 e^(-3).
a) The ODE is linear. In standard form we have p=2/t and g=(1/t^2)sin(t).
Using the integrating factor method we get the explicit solution:
y = [pi^2 y0/4 - cos(t)] / [t^2]. (Here "y0" is the initial condition)
This solution is defined in the interval (0,infty) about t0=pi/2.
b) Since p(t) and g(t) are continuous in (0,infty) about t0=pi/2,
the solution is unique by the Existence/Uniqueness Theorem for
linear IVPs.
c) For any y0 we get y(t) -> 0 as t -> infty.
a) The ODE can be separated as y^(-2)dy=t^(-1)dt. Integrating gives
the explicit solution:
y = y0 / [ 1 - y0 ln(t)], (where we are assuming t > 0 and y0 > 0)
To determine the interval of definition we start at t0 = 1.
We can go left just until t = 0, and we can go right just until
t = e^(1/y0) > 1. Thus, the interval of definition is
( 0, e^(1/y0) ).
b) Since f = (y^2)/t and df/dy = 2y/t are continuous at and around
( t0, y0 ) = ( 1 , y0 ) for any y0, the solution is unique by the
Existence/Uniqueness Theorem for general IVPs.
c) Our solution y(t) is defined only for t in ( 0, e^(1/y0) ). Thus,
we can not take the limit t -> infty.
a) F(y) = 2y^3 - 3y^2 - 2y = y(2y+1)(y-2) = 0 implies y=(-1/2), 0, 2.
Thus, there are three equilibrium solutions:
y1(t) = -1/2, .... y2(t)=0, .... y3(t)=2 .... defined in (-infty,infty)
b.1) The three equil solutions divide the (t,y) plane into
strips where F(y) is positive or negative. The solution
y(t) satisfying y(0)=1 is in a strip where F(y)<0, which
implies that this solution must be decreasing (y'<0) with
respect to t.
Thus y(t)>1 for some t>0 is *not* possible
since y(0)=1 and y(t) is decreasing for t>0.
b.2) Since F(y) and dF/dy(y) are continuous for all y, we have that
solution curves cannot intersect.
Thus y(t)<0 for some t>0 is *not* possible
since it cannot cross the equil solution y2(t)=0.