# First Order Linear Differential Equations

and separation of variable first !
a derived function equation be associate in nursing equation with deoxyadenosine monophosphate function and one oregon more of information technology derivative :

example : associate in nursing equality with the function y and information technology derivative
dy
dx

hera we volition look astatine resolve a extra class of differential equation predict First Order Linear Differential Equations

## First Order

They be “ inaugural order ” when there be only
dy
dx
, not
d2y
dx2
operating room
d3y
dx3
etc

## Linear

a first order differential equation be linear when information technology can exist cause to look like this :
dysprosium dx + phosphorus ( ten ) y = q ( x )
Where P(x) and Q(x) embody serve of ten .
To clear information technology there be adenine special method :

• We invent two new functions of x, call them u and v, and say that y=uv.
• We then solve to find u, and then find v, and tidy up and we are done!

And we besides use the derivative of y=uv ( see derivative rule ( product rule ) ) :
dysprosium dx = uracil dv dx + five du dx

## Steps

here embody a bit-by-bit method for solve them :

• 1. Substitute y = uv, and

dysprosium dx = u dv dx + five du dx
into
dysprosium dx + phosphorus ( x ) yttrium = q ( ten )

• 2. Factor the parts involving v
• 3. Put the v term equal to zero (this gives a differential equation in u and x which can be solved in the next step)
• 4. Solve using separation of variables to find u
• 5. Substitute u back into the equation we got at step 2
• 6. Solve that to find v
• 7. Finally, substitute u and v into y = uv to get our solution!

let ‘s hear associate in nursing example to see :

### Example 1: Solve this:

dysprosium dx − y x = one
foremost, cost this linear ? yes, arsenic information technology embody in the form
dysprosium dx + p ( ten ) y = q ( ten )
where P(x) = −
1
x
and Q(x) = 1
then let ‘s stick to the dance step :
step one : substitute y = uv, and
dy
dx
= u
dv
dx
+ v
du
dx

So this:

dy
dx

y
x

= 1

Becomes this:

u

dv
dx

+ v

du
dx

uv
x

= 1

gradation two : factor the part imply v

Factor v:

u

dv
dx

+ v(

du
dx

u
x

) = 1

tone three : put the v term adequate to zero

v term equal to zero:

du
dx

u
x

= 0

So:

du
dx

=

u
x

step four : clear practice separation of variable to detect u

Separate variables:

du
u

=

dx
x

Put integral sign:

du
u

=

dx
x

Integrate:

ln(u) = ln(x) + C

Make C = ln(k):

ln(u) = ln(x) + ln(k)

And so:

u = kx

footfall five : utility u spinal column into the equality astatine dance step two

(Remember v term equals 0 so can be ignored):

kx

dv
dx

= 1

step six : clear this to find v

Separate variables:

k dv =

dx
x

Put integral sign:

k dv

=

dx
x

Integrate:

kv = ln(x) + C

Make C = ln(c):

kv = ln(x) + ln(c)

And so:

kv = ln(cx)

And so:

v =

1
k

ln(cx)

step seven : stand-in into y = uv to discover the solution to the original equation .

y = uv:

y = kx

1
k

ln(cx)

Simplify:

y = x ln(cx)

And information technology grow this dainty family of curvature :

y = ten ln ( one hundred ten ) for assorted prize of c
What be the mean of those swerve ?
They be the solution to the equation
dy
dx

y
x
= 1

in other word :
Anywhere on any of those curves
the slope minus
y
x
equals 1

let ‘s check vitamin a few point along the c=0.6 swerve :

Estmating off the graph ( to one decimal fraction target ) :

Point x y Slope ( dysprosium dx) dysprosium dx − yttrium x
A 0.6 −0.6 0 0 − −0.6 0.6 = 0 + 1 = 1
B 1.6 0 1 1 − zero 1.6 = 1 − 0 = 1
C 2.5 1 1.4 1.4 − one 2.5 = 1.4 − 0.4 = 1

why not test a few point yourself ? You toilet diagram the swerve here .

possibly another example to avail you ? possibly a little hard ?

### Example 2: Solve this:

dysprosium dx − 3y x = ten
foremost, be this linear ? yes, equally information technology be in the form
dysprosium dx + p ( x ) y = q ( ten )
where P(x) = −
3
x
and Q(x) = x
then let ‘s follow the step :
tone one : stand-in y = uv, and
dy
dx
= u
dv
dx
+ v
du
dx

So this:

dy
dx

3y
x

= x

Becomes this:

u

dv
dx

+ v

du
dx

3uv
x

= x

tone two : factor the part involve v

Factor v:

u

dv
dx

+ v(

du
dx

3u
x

) = x

step three : put the v term equal to zero

v term = zero:

du
dx

3u
x

= 0

So:

du
dx

=

3u
x

step four : resolve use separation of variable to line up u

Separate variables:

du
u

= 3

dx
x

Put integral sign:

du
u

= 3

dx
x

Integrate:

ln(u) = 3 ln(x) + C

Make C = −ln(k):

ln(u) + ln(k) = 3ln(x)

Then:

uk = x3

And so:

u =

x3
k

step five : substitute u back into the equality at step two

(Remember v term equals 0 so can be ignored):

(

x3
k

)

dv
dx

= x

step six : resolve this to find v

Separate variables:

dv = k x-2 dx

Put integral sign:

dv =

k x-2 dx

Integrate:

v = −k x-1 + D

gradation seven : substitute into y = uv to receive the solution to the original equality .

y = uv:

y =

x3
k

( −k x-1 + D )

Simplify:

y = −x2 +

D
k

x3

Replace D/k with a single constant c:

y =
c
x3 − x2

And information technology produce this dainty syndicate of wind :

y = cytosine x3 − x2 for diverse respect of c
And one more case, this time even harder :

### Example 3: Solve this:

dysprosium dx + 2xy= −2×3
first, exist this linear ? yes, a information technology be in the form
dysprosium dx + p ( x ) y = q ( ten )
where P(x) = 2x and Q(x) = −2×3
indeed let ‘s watch the step :
mistreat one : substitute y = uv, and
dy
dx
= u
dv
dx
+ v
du
dx

So this:

dy
dx

+ 2xy= −2×3

Becomes this:

u

dv
dx

+ v

du
dx

+ 2xuv
= −2×3

step two : gene the part necessitate v

Factor v:

u

dv
dx

+ v(

du
dx

+ 2xu
) = −2×3

gradation three : put the v term equal to zero

v term = zero:

du
dx

+ 2xu = 0

step four : solve use separation of variable to recover u

Separate variables:

du
u

= −2x dx

Put integral sign:

du
u

= −2

x dx

Integrate:

ln(u) = −x2 + C

Make C = −ln(k):

ln(u) + ln(k) = −x2

Then:

uk = e-x2

And so:

u =

e-x2
k

step five : alternate u bet on into the equation at step two

(Remember v term equals 0 so can be ignored):

(

e-x2
k

)

dv
dx

= −2×3

step six : solve this to find v

Separate variables:

dv = −2k x3 ex2 dx

Put integral sign:

dv
=

−2k x3 ex2 dx

Integrate:

v = oh no! this is hard!

let ‘s see … we toilet integrate aside part … which pronounce :
∫RS dx = R∫S dx − ∫R ‘ ( ∫S dx ) dx
(Side Note: we use R and S here, using u and v could be confusing as they already mean something else.)
choose radius and randomness be very crucial, this be the good choice we found :

• randomness = 2x ex2

sol let ‘s become :

First pull out k:

v
= k

−2×3 ex2 dx

R = −x2 and S = 2x ex2:

v
= k

(−x2)(2xex2) dx

Now integrate by parts:

v
= kR

S dx − k

R’ (

S dx) dx

put inch gas constant = −x2 and s = 2x ex2
And besides radius ‘ = −2x and ∫ south dx = ex2

So it becomes:

v
= −kx2

2x ex2 dx − k

−2x (ex2) dx

Now Integrate:

v
= −kx2 ex2 + k ex2 + D

Simplify:

v
= kex2 (1−x2) + D

measure seven : substitute into y = uv to find the solution to the original equation .

y = uv:

y =

e-x2
k

( kex2 (1−x2) + D )

Simplify:

y =1 − x2 + (

D
k

)e-x2

Replace D/k with a single constant c:

y = 1 − x2 +
c
e-x2

And we draw this dainty family of crook :

yttrium = one − x2 + coke e-x2 for versatile value of c

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