# First Order Linear Differential Equations

You might wish to read about 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

, not

dy

dx

operating room

d2y

dx2

etc

d3y

dx3

## Linear

a

first order differential equationbelinearwhen information technology can exist cause to look like this :

dysprosiumdx+ phosphorus ( ten ) y = q ( x )

WhereP(x)andQ(x)embody serve of ten .

To clear information technology there be adenine special method :

- We invent two new functions of x, call them
uandv, and say thaty=uv.- We then solve to find
u, and then findv, and tidy up and we are done!And we besides use the derivative of

y=uv( see derivative rule ( product rule ) ) :

dysprosiumdx= uracil dvdx+ five dudx## Steps

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

- 1. Substitute
y = uv, anddysprosium

dx= u dvdx+ five dudx

into

dysprosiumdx+ phosphorus ( x ) yttrium = q ( ten )- 2. Factor the parts involving
v- 3. Put the
vterm equal to zero (this gives a differential equation inuandxwhich can be solved in the next step)- 4. Solve using separation of variables to find
u- 5. Substitute
uback into the equation we got at step 2- 6. Solve that to find
v- 7. Finally, substitute
uandvintoy = uvto get our solution!let ‘s hear associate in nursing example to see :

## Example 1: Solve this:

dysprosium

dx− yx= one

foremost, cost this linear ? yes, arsenic information technology embody in the form

dysprosiumdx+ p ( ten ) y = q ( ten )

whereP(x) = −and

1

x

Q(x) = 1

then let ‘s stick to the dance step :

step one : substitutey = 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

vFactor

v:u

dv

dx+ v(

du

dx−

u

x) = 1

tone three : put the

vterm adequate to zero

vterm equal to zero:

du

dx−

u

x= 0

So:

du

dx=

u

xstep four : clear practice separation of variable to detect

uSeparate variables:

du

u=

dx

xPut integral sign:

∫

du

u=

∫

dx

xIntegrate:

ln(u) = ln(x) + C

Make C = ln(k):

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

And so:

u = kx

footfall five : utility

uspinal column into the equality astatine dance step two(Remember

vterm equals 0 so can be ignored):kx

dv

dx= 1

step six : clear this to find

vSeparate variables:

k dv =

dx

xPut integral sign:

∫

k dv

=∫

dx

xIntegrate:

kv = ln(x) + C

Make C = ln(c):

kv = ln(x) + ln(c)

And so:

kv = ln(cx)

And so:

v =

1

kln(cx)

step seven : stand-in into

y = uvto discover the solution to the original equation .y = uv:

y = kx

1

kln(cx)

Simplify:

y = x ln(cx)

And information technology grow this dainty family of curvature :

y = ten ln ( one hundred ten ) for assorted prize ofc

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 thec=0.6swerve :

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

Point x y Slope ( dysprosium dx)dysprosium dx− yttriumxA 0.6 −0.6 0 0 − −0.6 0.6= 0 + 1 =1B 1.6 0 1 1 − zero 1.6= 1 − 0 =1C 2.5 1 1.4 1.4 − one 2.5= 1.4 − 0.4 =1why 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− 3yx= ten

foremost, be this linear ? yes, equally information technology be in the form

dysprosiumdx+ p ( x ) y = q ( ten )

whereP(x) = −and

3

x

Q(x) = x

then let ‘s follow the step :

tone one : stand-iny = 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

vFactor

v:u

dv

dx+ v(

du

dx−

3u

x) = x

step three : put the

vterm equal to zero

vterm = zero:

du

dx−

3u

x= 0

So:

du

dx=

3u

xstep four : resolve use separation of variable to line up

uSeparate variables:

du

u= 3

dx

xPut integral sign:

∫

du

u= 3

∫

dx

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Integrate:

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

Make C = −ln(k):

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

Then:

uk = x3

And so:

u =

x3

kstep five : substitute

uback into the equality at step two(Remember

vterm equals 0 so can be ignored):(

x3

k)

dv

dx= x

step six : resolve this to find

vSeparate 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 = uvto receive the solution to the original equality .y = uv:

y =

x3

k( −k x-1 + D )

Simplify:

y = −x2 +

D

kx3

Replace

D/kwith a single constantc:y =

c

x3 − x2And information technology produce this dainty syndicate of wind :

y = cytosine x3 − x2 for diverse respect ofc

And one more case, this time evenharder:## Example 3: Solve this:

dysprosium

dx+ 2xy= −2×3

first, exist this linear ? yes, a information technology be in the form

dysprosiumdx+ p ( x ) y = q ( ten )

whereP(x) = 2xandQ(x) = −2×3

indeed let ‘s watch the step :

mistreat one : substitutey = 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

= −2×3step two : gene the part necessitate

vFactor

v:u

dv

dx+ v(

du

dx+ 2xu

) = −2×3gradation three : put the

vterm equal to zero

vterm = zero:

du

dx+ 2xu = 0

step four : solve use separation of variable to recover

uSeparate 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

kstep five : alternate

ubet on into the equation at step two(Remember

vterm equals 0 so can be ignored):(

e-x2

k)

dv

dx= −2×3

step six : solve this to find

vSeparate 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 :

- radius = −x2 and
- randomness = 2x ex2
sol let ‘s become :

First pull out k:

v

= k∫

−2×3 ex2 dx

R = −x2andS = 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 + DSimplify:

v

= kex2 (1−x2) + Dmeasure seven : substitute into

y = uvto 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/kwith a single constantc:y = 1 − x2 +

c

e-x2And we draw this dainty family of crook :

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

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