First Order Differential Equations
BCCC Tutoring Center
Seperable Equations A differential equation is called seperable if it is of the form
g(y)y 0 = f (x)
An equation is seperable if we can isolate all y terms on one side of the equation and all x terms on
the other side. Equations of this type can be solved by integrating each side of the equation with
respect to the appropriate variable.
Examples
1. y 0 = yx
0
This equation is seperable, as can be seen after dividing by y. This gives yy = x. Integrating
both sides gives ln y = x + C =⇒ y = ex+C = Cex . When we divided by y, we tacitly
assumed that y 6= 0. We must therefore check if y = 0 solves the differential equation. The
soltuions are then y = 0 and y = Cex .
2. 2xy 2 − x4 y 0 = 0
We can rearrange this equation to give
by integrating.
−1
x2
+C =
−1
y
=⇒ y =
2
x3
=
y0
.
y2
x2
1+Cx2
This is seperable, and the solution is revealed
.
First Order Linear Equations These differential equations take the general form
y 0 + p(x)y = q(x)
where p(x) and q(x) are functions of x only. The following are examples of linear equations.
1. y 0 + x2 y = 0
2. y 0 + cos(x) y = x2
3. y 0 +
y
1−x
= ex
The following equations would not qualify as linear.
1. (y 0 )2 − sin(x) y = 0
2. y 0 +
x2
y
= 2x
3. y 0 + ex y = y 2
1
To solve these equations, we use the integrating
factor µ = e
R
1
factor, the solution can then be written as y = µ µ q(x) dx.
R
p(x) dx
. With this integrating
Examples
1. y 0 +
y
x
= 2ex
2
R
1
In this case, p(x) = x1 and µ = e x dx = eln x = x. Using our above equation for y gives the
R
2
2
solution y = x1 2xex dx = x1 (ex + C)
2. y 0 + y cos x = cos x
R
In this case, p(x)
e cos x dx = esin x . Again, applying the solution equation
R = cossinxx and µ =
1
− sin x sin x
dx = e
(e
+ C) = 1 + Ce− sin x
gives y = esin x cos x e
Exact Equations An equation of the form
M dx + N dy = 0
with M and N functions of x and y, is said to be exact if
∂M
∂y
=
∂N
.
∂x
To solve an exact equation, we follow these steps:
R
1. Our solution will be F (x, y) = Ψ(y) + M dx = C, where Ψ(y) is a function entirely of y to
be found later.
R
2. Calculate the integral M dx.
3. Take the derivative
of F (x, y) with respect to y. Set this equal to N and solve for Ψ0 (y).
R
∂
M
dx
Ψ0 (y) = N − ∂y .
R
4. Find Ψ(y) by integrating Ψ0 (y) with respect to y. Ψ(y) = Ψ0 (y) dy.
5. Plug Ψ(y) into F (x, y) to obtain the solution.
Examples
1. 2xy dx + (x2 + 2y) dy = 0
Here M = 2xy and N = x2 + 2y. We see the equation is exact since ∂M
= 2x = ∂N
.
∂y
∂x
R
2 y)
∂(x
F (x, y) = 2xy dx + Ψ(y) = x2 y + Ψ(y). Now we solve for Ψ(y). Ψ0 (y) = N − ∂y =
(x2 + 2y) − x2 =⇒ Ψ0 (y) = 2y. Integrating we see that Ψ(y) = y 2 . Our solution is then
x2 y + y 2 = c.
2
2. (2xy − 9x2 ) dx + (2y + x2 + 1) dy = 0
=
Here M = 2xy − 9x2 and N = 2y + x2 + 1. We see the equation is exact since ∂M
∂y
R
∂N
2
2
3
2x = ∂x . F (x, y) = 2xy − 9x dx + Ψ(y) = x y − 3x + Ψ(y). Next, solve for Ψ(y).
2
3)
Ψ0 (y) = N − ∂(x y−3x
= (2y + x2 + 1) − x2 = 2y + 1. Integrate this to see that Ψ(y) = y 2 + y.
∂y
The solution is then F (x, y) = x2 y − 3x3 + y 2 + y = C.
Making Equations Exact Ocassionally, one will encounter an equation of the form
M dx + N dy = 0
that does not meet the criterion for exactness. In certain situations, we can find an appropriate
integrating factor which will transform this into an exact equation.
Case 1 Integrating factors of x only: If the quantity p(x) =
occurances of y, then µ = e
R
p(x) dx
occurances of x, then µ = e
p(y) dy
−
∂N
∂x
∂N
∂x
−
∂M
∂y
is a function with no
N
is an integrating factor for the differential equation.
is a function with no
M
is an integrating factor for the differential equation.
Case 2 Integrating factors of y only: If the quantity p(y) =
R
∂M
∂y
When the integrating factor µ exists, one may multiply the differential equation by µ to created
an exact equation.
Examples
1. (y 2 (x2 + 1) + xy) dx + (2xy + 1) dy = 0
∂M
∂y
= 2y(x2 + 1) + x, and
for an integrating factor.
R
x so that µ = e
x dx
∂N
∂x
∂M
∂y
x2
=ex
= 2y. As we can see, this equation is not exact. We will search
− ∂N
2 +1)+x−2y
2 +x
∂x
= 2y(x 2yx+1
= 2yx
= x. This a function entirely of
2yx+1
N
will be an integrating factor.
x2
x2
x2
x2
Multiply the initial equation by µ to give (e 2 y 2 (x2 + 1) + e 2 xy) dx + (2e 2 xy + e 2 ) dy = 0.
x2
x2
x2
Now ∂M
= 2x2 e 2 y + 2ye 2 + xe 2 = ∂N
so that the equation is now exact and can be solved
∂y
∂x
via the methods previously discussed.
2. (x2 y + 2y 2 sin x) dx + ( 23 x3 − 6y cos x) dy = 0
The equation is not exact since ∂M
= x2 + 4y sin x, and ∂N
= 2x3 + 6y sin x. Now attempt to
∂y
∂x
∂N
− ∂M
2x2 + 6y sin x − x2 − 4y sin x
x2 + 2y sin x
∂x
∂y
find an integrating factor.
=
=
= y1 .
M
x2 y + 2y 2 sin x
x2 y + 2y 2 sin x
3
This is a function entirely of y so the equation has an integrating factor of the form e
eln y = y.
R
1
y
dy
=
Multiply the initial equation by y to give (x2 y 2 + 2y 3 sin x) dx + ( 32 x3 y − 6y 2 cos x) dy = 0.
Now ∂M
= 2x2 y + 6y 2 sin x = ∂N
. As we can see, this equation is now exact and can be solved
∂y
∂x
accordingly.
4