Differential Equations
Solving First-Order Linear DEs
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A First-Order Linear Differential Equation
can always be put into the form:
y  p(t)  y  f(t)
The general solution will always have the form:
y(t)  yh  yp
yh is the solution to the corresponding
homogeneous equation, where f(t)=0
yp is a particular solution to the original DE.
We will have four main ways to solve this type of equation.
One of them, separation of variables, has already been covered.
The other methods are:
*Integrating Factor Method
*Variation of Parameters (Euler-Lagrange Method)
*Undetermined Coefficients (i.e. guess-and-check method)Prepared by Vince Zaccone
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Here are a few examples of linear, first order DEs.
Put each of them in the standard format:
y  p(t)  y  f(t)
1)
y  t  y  6t
2)
t  y  y  sin(t)
3)
y  3y  2e5t
4)
t  y  4y  t2
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Here are a few examples of linear, first order DEs.
Put each of them in the standard format:
y  p(t)  y  f(t)
1)
y  t  y  6t
2)
t  y  y  sin(t)
3)
y  3y  2e5t
4)
t  y  4y  t2
y  t  y  6t
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Here are a few examples of linear, first order DEs.
Put each of them in the standard format:
y  p(t)  y  f(t)
y  t  y  6t
y  t  y  6t
2)
t  y  y  sin(t)
1
sin(t)
y  y 
t
t
3)
y  3y  2e5t
4)
t  y  4y  t2
1)
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Here are a few examples of linear, first order DEs.
Put each of them in the standard format:
y  p(t)  y  f(t)
y  t  y  6t
y  t  y  6t
2)
t  y  y  sin(t)
1
sin(t)
y  y 
t
t
3)
y  3y  2e5t
y  3y  2e5t
4)
t  y  4y  t2
1)
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Here are a few examples of linear, first order DEs.
Put each of them in the standard format:
y  p(t)  y  f(t)
y  t  y  6t
y  t  y  6t
2)
t  y  y  sin(t)
1
sin(t)
y  y 
t
t
3)
y  3y  2e5t
y  3y  2e5t
4)
2
1)
t  y  4y  t
4
y  y  t
t
Now that they are in the right format let’s try solving them.
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1)
y  t  y  6t
Find the general solution to this DE.
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1)
y  t  y  6t
Find the general solution to this DE.
We have several options for this one. The equation is not separable,
but we can use an integrating factor, or variation of parameters.
Let’s try the integrating factor.
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1)
y  t  y  6t
Find the general solution to this DE.
We have several options for this one. The equation is not separable,
but we can use an integrating factor, or variation of parameters.
Let’s try the integrating factor.
First we have to calculate our factor [call it µ(t)].
Here is the formula:
(t)  e p(t)dt
So we have to integrate the function in front of the y in the equation.
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1)
y  t  y  6t
Find the general solution to this DE.
We have several options for this one. The equation is not separable,
but we can use an integrating factor, or variation of parameters.
Let’s try the integrating factor.
First we have to calculate our factor [call it µ(t)].
Here is the formula:
(t)  e p(t)dt
So we have to integrate the function in front of the y in the equation.
1
 tdt   2 t
(t)  e
2
Don’t worry about the constant of
integration here, it won’t matter.
2
1
t
2
Once we have this integrating factor, the trick is to multiply it
through the original equation, and then stare at the left-hand side.
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1)
y  t  y  6t
Find the general solution to this DE.
Here is the equation after we multiply through by the integrating factor:
e
2
1
t
2
 y  te
2
1
t
2
 y  6te
2
1
t
2
Look at the left-hand side and try to figure out the trick.
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1)
y  t  y  6t
Find the general solution to this DE.
Here is the equation after we multiply through by the integrating factor:
e
2
1
t
2
 y  te
2
1
t
2
 y  6te
2
1
t
2
The left-hand side will always be the derivative of a product. Apply the
product rule to the left side and prove it to yourself:

e

2
1
t
2

2
1
t

 y  6te 2

At this point we simply have to realize that if we integrate both sides,
we can then divide by the integrating factor and we have it solved.
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1)
y  t  y  6t
Find the general solution to this DE.
We will need to integrate the right side (the left side is already done)
e
2
1
t
2
 y   6te
2
1
t
2
dt
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1)
y  t  y  6t
Find the general solution to this DE.
We will need to integrate the right side (the left side is already done)
e
2
1
t
2
 y   6te
2
1
t
2
dt
Integrate by substitution:
 6te
2
1
t
2
dt
u   12 t2
   6eudu
 6e
2
1
t
2
du  tdt
C
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1)
y  t  y  6t
Find the general solution to this DE.
We will need to integrate the right side (the left side is already done)
e
e
2
1
t
2
2
1
t
2
 y   6te
 y  6e
2
1
t
2
2
1
t
2
dt
C
Integrate by substitution:
 6te
2
1
t
2
dt
u   12 t2
   6eudu
 6e
2
1
t
2
du  tdt
C
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1)
y  t  y  6t
Find the general solution to this DE.
We will need to integrate the right side (the left side is already done)
e
e
2
1
t
2
2
1
t
2
 y   6te
 y  6e
2
1
t
2
2
1
t
2
dt
C
Integrate by substitution:
 6te
2
1
t
2
dt
u   12 t2
   6eudu
Now divide to solve for y:
y(t)  6  Ce
1 t2
2
 6e
2
1
t
2
du  tdt
C
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1)
y  t  y  6t
Find the general solution to this DE.
We will need to integrate the right side (the left side is already done)
e
e
2
1
t
2
2
1
t
2
 y   6te
 y  6e
2
1
t
2
2
1
t
2
dt
C
Integrate by substitution:
 6te
2
1
t
2
dt
u   12 t2
   6eudu
Now divide to solve for y:
1 t2
2
y(t)  
6  Ce


yp
 6e
2
1
t
2
du  tdt
C
yh
This is the general solution. Notice that it has 2 parts added together.
These are the homogeneous and particular solutions that we expected.
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Integrating Factor Method
Here is a summary of the method:
Put the Linear, First-Order DE in standard form.
Next, calculate the Integrating Factor.
y  p(t)  y  f(t)
(t)  e p(t)dt
You can multiply through the DE. It should look
like this:
(t)  y  (t)  p(t)  y  (t)  f(t)
Now you recognize that the left side IS the derivative of µy.
Integration yields:
This constant of integration is super-
important. It is where the homogeneous
part of the solution comes from.
(t)  y   (t)  f(t)  dt  C
1
C
y(t) 
 (t)  f(t)  dt 
(t)
(t)
 
yp
You can go straight to this
final formula if you like.
yh
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2) y 
1
sin(t)
y
t
t
Find the general solution to this DE.
Again we have several options for this one. This time let’s use the
Euler-Lagrange method, also called “variation of parameters”.
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2) y 
1
sin(t)
y
t
t
Find the general solution to this DE.
Again we have several options for this one. This time let’s use the
Euler-Lagrange method, also called “variation of parameters”.
This method is done in two stages. First we will solve the
homogeneous version of the DE. Then we will use that solution
to manufacture a particular solution to the original equation.
Write down the associated homogeneous DE:
y 
1
y0
t
This will be separable. Always.
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2) y 
1
sin(t)
y
t
t
Find the general solution to this DE.
Again we have several options for this one. This time let’s use the
Euler-Lagrange method, also called “variation of parameters”.
This method is done in two stages. First we will solve the
homogeneous version of the DE. Then we will use that solution
to manufacture a particular solution to the original equation.
Write down the associated homogeneous DE:
y 
1
y0
t
This will be separable. Always.
dy
1
dy
1
 y
  dt
dt
t
y
t
dy
1
   dt

y
t
ln y   ln t  C
yh  e
ln t  C

C
t
Here is the homogeneous solution.
It will have an arbitrary constant.
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2) y 
1
sin(t)
y
t
t
Find the general solution to this DE.
Now for stage two. We will use the homogeneous solution as a
model for a particular solution. The trick is that instead of an
arbitrary constant, we assume that there is an arbitrary function.
v(t)
yp 
t
We will substitute this into the original DE
to make a new equation that we can solve
for our new function v(t).
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2) y 
1
sin(t)
y
t
t
Find the general solution to this DE.
Now for stage two. We will use the homogeneous solution as a
model for a particular solution. The trick is that instead of an
arbitrary constant, we assume that there is an arbitrary function.
v(t)
yp 
t
yp 
We will substitute this into the original DE
to make a new equation that we can solve
for our new function v(t).
v  t  v  1
Quotient rule
t2
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2) y 
1
sin(t)
y
t
t
Find the general solution to this DE.
Now for stage two. We will use the homogeneous solution as a
model for a particular solution. The trick is that instead of an
arbitrary constant, we assume that there is an arbitrary function.
v(t)
yp 
t
yp 
We will substitute this into the original DE
to make a new equation that we can solve
for our new function v(t).
v  t  v  1
Quotient rule
t2
vt  v 1 v sin(t)


2
t t
t
t
v sin(t)
There should always be some nice cancellation here.

t
t
v  sin(t)
v   cos(t)
We don’t need the constant of integration here.
It would be redundant since we already have the
homogeneous solution.
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2) y 
1
sin(t)
y
t
t
Find the general solution to this DE.
We are finally ready to put together our solution.
C
This is what we found in stage 1.
t
v(t)  cos(t)
yp 

t
t
yh 
ygeneral 
C cos(t)

t
t
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Euler-Lagrange Method - Variation of Parameters
Here is a summary of the method:
Write the Linear, First-Order DE in standard form:
y  p(t)  y  f(t)
Solve the corresponding homogeneous equation:
y  p(t)  y  0  yh  Ce  p(t)dt
The particular solution will be the same as the homogeneous, but with
the arbitrary CONSTANT replaced by an arbitrary FUNCTION of t.
yp  v(t)  e  p(t)dt
After taking the derivative and plugging into the original DE, you get
a differential equation for v(t). The solution will be:
v(t)   f(t)  e p(t)dtdt
Adding the homogeneous and particular solutions yields the answer:
  p(t)dt
  p(t)dt
 p(t)dt
y(t)  Ce

e

f
(
t
)

e
dt
 
 


yh
yp
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3)
y  3y  2e5t
Find the general solution to this DE.
For this one we can start by finding the homogeneous solution, as if we are going to use
variation of parameters. Then, instead of continuing with that method, we will make an
educated guess to find the particular solution.
First, the homogeneous equation:
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3)
y  3y  2e5t
Find the general solution to this DE.
For this one we can start by finding the homogeneous solution, as if we are going to use
variation of parameters. Then, instead of continuing with that method, we will make an
educated guess to find the particular solution.
First, the homogeneous equation:
y  3y  0  yh  Ce3t
This should be a familiar solution by now.
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3)
y  3y  2e5t
Find the general solution to this DE.
For this one we can start by finding the homogeneous solution, as if we are going to use
variation of parameters. Then, instead of continuing with that method, we will make an
educated guess to find the particular solution.
First, the homogeneous equation:
y  3y  0  yh  Ce3t
This should be a familiar solution by now.
Look at the right-hand-side of the original DE and try to guess what the particular
solution should be.
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3)
y  3y  2e5t
Find the general solution to this DE.
For this one we can start by finding the homogeneous solution, as if we are going to use
variation of parameters. Then, instead of continuing with that method, we will make an
educated guess to find the particular solution.
First, the homogeneous equation:
y  3y  0  yh  Ce3t
This should be a familiar solution by now.
Look at the right-hand-side of the original DE and try to guess what the particular
solution should be. We should try something like:
yp  Ae5t
A good first guess often uses something that looks
like the right side, but with arbitrary constants.
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3)
y  3y  2e5t
Find the general solution to this DE.
For this one we can start by finding the homogeneous solution, as if we are going to use
variation of parameters. Then, instead of continuing with that method, we will make an
educated guess to find the particular solution.
First, the homogeneous equation:
y  3y  0  yh  Ce3t
This should be a familiar solution by now.
Look at the right-hand-side of the original DE and try to guess what the particular
solution should be. We should try something like:
yp  Ae5t
A good first guess often uses something that looks
like the right side, but with arbitrary constants.
Plug this into the DE and you should get an equation for the arbitrary constant A:
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3)
y  3y  2e5t
Find the general solution to this DE.
For this one we can start by finding the homogeneous solution, as if we are going to use
variation of parameters. Then, instead of continuing with that method, we will make an
educated guess to find the particular solution.
First, the homogeneous equation:
y  3y  0  yh  Ce3t
This should be a familiar solution by now.
Look at the right-hand-side of the original DE and try to guess what the particular
solution should be. We should try something like:
yp  Ae5t
A good first guess often uses something that looks
like the right side, but with arbitrary constants.
Plug this into the DE and you should get an equation for the arbitrary constant A:
y  3y  2e5t
5Ae5t  3Ae5t  2e5t
2A  2  A  1
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3)
y  3y  2e5t
Find the general solution to this DE.
For this one we can start by finding the homogeneous solution, as if we are going to use
variation of parameters. Then, instead of continuing with that method, we will make an
educated guess to find the particular solution.
First, the homogeneous equation:
y  3y  0  yh  Ce3t
This should be a familiar solution by now.
Look at the right-hand-side of the original DE and try to guess what the particular
solution should be. We should try something like:
yp  Ae5t
A good first guess often uses something that looks
like the right side, but with arbitrary constants.
Plug this into the DE and you should get an equation for the arbitrary constant A:
y  3y  2e5t
Now we have our answer:
5Ae5t  3Ae5t  2e5t
2A  2  A  1
3t
5t
y(t)  Ce

e


 
yh
yp
This method of “undetermined coefficients” will be used
extensively for 2nd-order DEs in Math 5A.
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4)
t  y  4y  t2
Find the general solution to this DE.
Now that we have all the methods, try this one on your own.
Use any method you like.
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4)
t  y  4y  t2
Find the general solution to this DE.
Now that we have all the methods, try this one on your own.
Use any method you like.
The answer is:
y(t)  Ct 4  16 t2
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