# 4. Linear DEs of Order 1

If *P = P*(*x*) and *Q = Q*(*x*)
are functions of *x* only, then

`(dy)/(dx)+Py=Q`

is called a **linear differential equation order 1.**

We can solve these **linear** DEs using an **integrating
factor****.**

For linear DEs of order 1, the integrating factor is:

`e^(int P dx`

The **solution** for the DE is given by multiplying *y* by the integrating factor (on the left) and multiplying *Q *by the integrating factor (on the right) and integrating the right side with respect to *x*, as follows:

`ye^(intP dx)=int(Qe^(intP dx))dx+K`

### Example 1

Solve `(dy)/(dx)-3/xy=7`

Answer

`(dy)/(dx)-3/xy=7`

Here, `P(x)=-3/x` and `Q(x) = 7`.

Now for the integrating factor:

`"IF"=e^(intPdx)` `=e^(int-3/xdx` `=e^(-3 ln x)` `=e^(ln x^-3)` `=x^-3`

[If you are not sure what just happened, check out Integration: Basic Logarithm Form and Logarithms to Base *e*.]

We need to apply the following formula: `ye^(intPdx)=int(Qe^(intPdx))dx`

For the left hand side of the formula, we have

`ye^(intPdx) = yx^-3`

For the right hand of the formula, *Q* = 7 and the IF = *x*^{-3}, so:

`Qe^(intPdx)=7x^-3`

Applying the outer integral:

`int(Qe^(intPdx))dx=int7x^-3dx =` `-7/2x^-2+K`

Now, applying the whole formula, `ye^(intPdx)=intQe^(intPdx)dx`, we have

`yx^-3=-7/2x^-2+K`

Multiplying throughout by *x*^{3} gives the general solution for `y`.:

`y=-7/2x+Kx^3`

Is it correct? Differentiate this answer to make sure it produces the differential equation in the question.

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Here is the solution graph of our answer for Example 1 (I've used *K* = 0.5).

It is a cubic polynomial curve:

Solution using `K=0.5`.

### Example 2

Solve `(dy)/(dx)+(cot x)y=cos x`

Answer

`(dy)/(dx)+(cot x)y=cos x`

Here, `P(x)=cot x` and `Q(x)=cos x`.

`"IF"=e^(intPdx)` `=e^(int cot x dx)` `=e^(ln sin x)` `=sin x`

(See Integration of Trigonometric Forms for an explanation of `int cot x dx`.)

Now `Qe^(intPdx)=cos x sin x`.

Apply the formula, `ye^(intPdx)=intQe^(intPdx)dx` to obtain:

`y sin x=intcos x sin x dx`

The integral needs a simple substitution: `u = sin x`, `du = cos x dx`

`y sin x=(sin^2x)/2+K`

Divide throughout by sin *x* to get the required general solution of the DE:

`y=(sin x)/2+K csc x`

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Here is the solution graph of our answer for Example 2 (I've used *K* = 0.1) .

It is a composite trigonometric curve, where the main shape is the cosecant curve, and the "wiggles" are due to the addition of the (sin *x*)/2 part:

Typical solution graph using `K=0.1`.

### Example 3

Solve `dy + 3y dx = e^(-3x)dx`

Answer

Dividing throughout by *dx* to get the equation in the required form, we get:

`(dy)/(dx)+3y=e^(-3x)`

In this example, *P*(*x*) = 3 and *Q*(*x*) = *e*^{−3x}.

Now the integrating factor in this example is

`e^(intPdx)=e^(int3dx)=e^(3x)`

and

`intQe^(intPdx)dx=inte^(-3x)e^(3x)dx`

`=int1 dx`

`=x`

Using `ye^(intPdx)=intQe^(intPdx)dx+K`, we have:

ye^{3x}=x + K

or we could write it as:

`y=(x+K)/e^(3x)`

Here is the solution graph for Example 3 (I've used *K* = 5).

It was necessary to zoom out (a lot) to see what is going on in this graph.

Typical solution graph using `K=5`.

### Example 4

Solve `2(y - 4x^2)dx + x dy = 0`

Answer

We need to get the equation in the form of a linear DE of order 1.

Expand the bracket and divide throughout by *dx*:

`2y-8x^2+x(dy)/(dx)=0`

Rearrange:

`x(dy)/(dx)+2y=8x^2`

Divide throughout by *x*:

`(dy)/(dx)+2/xy=8x`

Here, `P(x)=2/x` and `Q(x) = 8x`.

`"IF"=e^(intPdx)` `=e^(int2/xdx)` `=e^(2 ln x)` `=e^(ln x^2)` `=x^2`

Now

`Qe^(intPdx)=(8x) x^2=8x^3`

Applying the formula:

`ye^(intPdx)=intQe^(intPdx)dx+K`

gives:

`y\ x^2=int8x^3dx+K=2x^4+K`

Divide throughout by *x*^{2}:

`y=2x^2+K/x^2`

#### Solving directly, using Scientific Notebook

Scientific Notebook cannot solve our original question:

`2(y-4x^2)dx+x\ dy=0`

We have to rearrange it in terms of `(dy)/(dx)` and then solve it using

**Compute** menu → **Solve ODE...** → **Exact**

Easy to understand math videos:

MathTutorDVD.com

Here is the solution graph for Example 4 (I've used *K* = 5).

There is a discontitnuity at *x* = 0.

Typical solution graph using `K=5`.

### Example 5

Solve `x(dy)/(dx)-4y=x^6e^x`

Answer

Divide throughout by *x*:

`(dy)/(dx)-4/xy=x^5e^x`

Here,

`P(x)=-4/x` and `Q(x)=x^5e^x`.

`"IF"=e^(intPdx)` `=e^(int-4/xdx)` `=e^(-4ln\ x)` `=e^(ln\ x^-4)` `=x^-4`

Now

`Qe^(intPdx)=(x^5e^x) x^-4=xe^x`

Applying the formula: `ye^(intPdx)=intQe^(intPdx)dx+K` gives

`y\ x^(-4)=intxe^xdx+K`

This requires integration by parts, with

`u=x,` and `dv=e^xdx`

This gives us

`du=dx,` and `v=e^x`.

So

`y\ x^-4=xe^x-e^x+K`

Multiplying throughout by *x*^{4} gives us `y` as an explicit function of `x`:

`y=x^5e^x-x^4e^x+Kx^4`

Here is the solution graph for Example 5 (I've used *K* = 0.005).

Typical solution graph using `K=0.005`.

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