Consider the Differential Equation Dy/dx=x/y Where Y Cannot Equal 0

First Order Linear Differential Equations

You might like to scan about Differential Equations
and Legal separation of Variables starting time!

A Differential Equation is an equation with a occasion and cardinal or more of its derivatives:

Example: an equivalence with the serve y and its derived function dy dx

Here we will look at resolution a special class of Differential Equations called Archetypical Order Linear Differential Equations

Eldest Order

They are "First Order" when there is only dy dx , not d²y dx² or d³y dx³ etc

Linear

A first order differential equation is lengthways when it can atomic number 4 made to look on the likes of this:

Dy dx + P(x)y = Q(x)

Where P(x) and Q(x) are functions of x.

To solve information technology at that place is a specialized method:

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

And we likewise use the derivative of y=uv (see Derivative instrument Rules (Ware Rule) ):

dy dx = u dv dx + v du dx

Stairs

Here is a abuse-by-step method acting for solving them:

• 1. Substitute y = uv, and

  dy dx = u dv dx + v du dx

  into

  dy dx + P(x)y = Q(x)

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

Let's try an example to see:

Example 1: Solve this:

dy dx − y x = 1

First, is this analogue? Yes, as it is in the form

dysprosium dx + P(x)y = Q(x)
where P(x) = − 1 x and Q(x) = 1

So let's follow the stairs:

Step 1: Substitute y = UV, and dysprosium dx = u dv dx + v du dx

Thusly this: dy dx − y x = 1

Becomes this: u dv dx + v du dx − uv x = 1

Mistreat 2: Factor the parts involving v

Factor v: u dv dx + v( du dx − u x ) = 1

Step 3: Put together the v term fifty-fifty to cipher

v term equal to zero: du dx − u x = 0

So: du dx = u x

Step 4: Solve victimization separation of variables to find u

Separate variables: du u = dx x

Frame integral sign: ∫ du u = ∫ dx x

Integrate: ln(u) = ln(x) + C

Make C = ln(k): ln(u) = ln(x) + ln(k)

And soh: u = kx

Step 5: Substitute u back into the equation at Abuse 2

(Remember v term equals 0 so can be ignored): kx dv dx = 1

Step 6: Solve this to find v

Separate variables: k dv = dx x

Redact integral sign: ∫k dv = ∫ dx x

Integrate: kv = ln(x) + C

Make C = ln(c): kv = ln(x) + ln(c)

And soh: kv = ln(cx)

And and then: v = 1 k ln(cx)

Step 7: Substitute into y = uv to find the solution to the original equation.

y = uv: y = kx 1 k ln(cx)

Simplify: y = x ln(110)

And information technology produces this nice family of curves:

y = x ln(cx) for various values of c

What is the substance of those curves?

They are the solution to the equation dy dx − y x = 1

In other words:

Anywhere on any of those curves
the slope disadvantageous y x equals 1

Let's check a few points on the c=0.6 curve:

Estmating off the graph (to 1 decimal place):

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

Why non test a couple of points yourself? You can plot the curve here.

Perhaps other instance to help you? Maybe a lilliputian harder?

Example 2: Solve this:

dysprosium dx − 3y x = x

First, is this linear? Yes, as IT is in the mannequin

atomic number 66 dx + P(x)y = Q(x)
where P(x) = − 3 x and Q(x) = x

And so let's follow the steps:

Step 1: Ersatz 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

Stride 2: Factor in the parts involving v

Factor v: u dv dx + v( du dx − 3u x ) = x

Step 3: Invest the v term equal to zero

v term = nought: du dx − 3u x = 0

So: du dx = 3u x

Measure 4: Solve using legal separation of variables to happen u

Separate variables: du u = 3 dx x

Put intrinsic sign up: ∫ du u = 3 ∫ dx x

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

Make C = −ln(k): ln(u) + ln(k) = 3ln(x)

Then: GB = x³

And so: u = x³ k

Step 5: Substitute u back into the equation at Step 2

(Remember v term equals 0 so can be ignored): ( x³ k ) dv dx = x

Step 6: Solve this to find v

Discrete variables: dv = k x^-2 dx

Put inherent sign: ∫dv = ∫k x^-2 dx

Integrate: v = −k x^-1 + D

Step 7: Substitute into y = uv to find the solution to the innovational equation.

y = ultraviolet radiation: 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 it produces this nice family of curves:

y = c x³ − x² for assorted values of c

And one more example, this time even harder:

Example 3: Wor this:

dy dx + 2xy= −2x³

Early, is this lineal? Yes, every bit information technology is in the form

dy dx + P(x)y = Q(x)
where P(x) = 2x and Q(x) = −2x³

And so countenance's follow the stairs:

Step 1: Substitute y = uv, and Dy dx = u dv dx + v du dx

So this: dy dx + 2xy= −2x³

Becomes this:  u dv dx + v du dx + 2xuv = −2x³

Footmark 2: Factor the parts involving v

Factor v: u dv dx + v( du dx + 2xu ) = −2x³

Step 3: Put the v term equal to zero

v terminus = zero: du dx + 2xu = 0

Step 4: Wor using interval of variables to find u

Secern 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

Abuse 5: Substitute u back into the equation at Step 2

(Remember v term equals 0 so can be unnoticed): ( e^-x2 k ) dv dx = −2x³

Step out 6: Solve this to find v

Separate variables: dv = −2k x³ e^x2 dx

Put integral sign: ∫dv = ∫−2k x³ e^x2 dx

Integrate: v = oh no! this is hard!

Let's see ... we can integrate by parts... which says:

∫RS dx = R∫S dx − ∫R' ( ∫S dx ) dx

(Side Bill: we employment R and S here, exploitation u and v could be confusing as they already poor something else.)

Choosing R and S is very important, this is the best choice we constitute:

• R = −x² and
• S = 2x e^x2

Soh let's go:

First pull out k: v = k∫−2x³ e^x2 dx

R = −x² and S = 2x e^x2 : v = k∫(−x²)(2xe^x2 ) dx

Now integrate by parts: v = kR∫S dx − k∫R' ( ∫ S dx) dx

Barge in R = −x² and S = 2x e^x2

And also R' = −2x and ∫ S dx = e^x2

Thusly information technology becomes: v = −kx² ∫2x e^x2 dx − k∫−2x (e^x2 ) dx

Now Integrate: v = −kx² e^x2 + k e^x2 + D

Simplify: v = ke^x2 (1−x²) + D

Step 7: Relief into y = ultraviolet light to line up the solution to the original equation.

y = uv: y = e^-x2 k ( ke^x2 (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 get this nice family of curves:

y = 1 − x² + c e^{- x2} for various values of c

9429, 9430, 9431, 9432, 9433, 9434, 9435, 9436, 9437, 9438

Consider the Differential Equation Dy/dx=x/y Where Y Cannot Equal 0

Source: https://www.mathsisfun.com/calculus/differential-equations-first-order-linear.html

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