3.1. Derivative of the first function over y is −2x. Derivative of the second
function over x is −2x, hence the equation is exact.
We want to find a function P (x, y) such that
∂P
= 3x2 − 2xy + 2
∂x
∂P
= 6y 2 − x2 + 3
∂y
We get from the first equation that P (x, y) = x3 −x2 y +2x+C(y). Substituting
this into the second equation we get
−x2 + C 0 (y) = 6y 2 − x2 + 3,
C 0 (y) = 6y 2 + 3,
hence we can take C(y) = 2y 3 + 3y, and P (x, y) = x3 − x2 y + 2x + 2y 3 + 3y, so
that the general solution of the equation is
x3 − x2 y + 2x + 2y 3 + 3y = C.
3.2. The equation is exact if
∂(x + y)x2
∂(xy 2 + bx2 y)
=
,
∂y
∂x
that is
2xy + bx2 = 3x2 + 2xy,
or b = 3. If b = 3, then in order to solve the equation we have to find P (x, y)
such that
∂P
= xy 2 + 3x2 y
∂x
∂P
= x3 + x2 y
∂y
From the first equality, we have
P (x, y) =
x2 y 2
+ x3 y + C(y),
2
hence from the second equality, we get
x2 y + x3 + C 0 (y) = x3 + x2 y,
i.e., we can take C(y) = 0, and the general solution is
x2 y 2
+ x3 y = C.
2
3.3. The characteristic polynomial is 6λ2 − λ − 1 = 0, hence the roots are
λ1 = − 31 , and λ2 = 1/2. It follows that the general solution of the differential
equation is
y(t) = C1 e−t/3 + C2 et/2 .
1
3.4. The characteristic polynomial is λ2 − 1 = 0, hence the general solution
is y(t) = C1 et + C2 e−t . Then y 0 (t) = C1 et − C2 e−t , so that the initial condition
gives the system
C1 + C2 = 45
C1 − C2 = − 43
whence we have C1 =
1
4
and C2 = 1, so that the solution is
y(t) =
3.5.
2t
e
2e2t
1 t
e + e−t .
4
3
7
e−3t/2 = − et/2 − 2et/2 = − et/2 .
− 32 e−3t/2 2
2
2