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4. Linear DEs of Order 1

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


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`



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:


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


Multiplying throughout by x3 gives the general solution for `y`.:


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

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

It is a cubic polynomial curve:

solution of differential equation - cubic

Solution using `K=0.5`.

Example 2

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


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:

solution of differential equation - composite trigonometric curve

Typical solution graph using `K=0.1`.

Example 3

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


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


In this example, P(x) = 3 and Q(x) = e−3x.

Now the integrating factor in this example is




`=int1 dx`


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

ye3x = x + K

or we could write it as:


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.

solution of differential equation - zoom out

Typical solution graph using `K=5`.

Example 4

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


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

Expand the bracket and divide throughout by dx:




Divide throughout by x:


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`


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

Applying the formula:



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

Divide throughout by x2:


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

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

There is a discontitnuity at x = 0.

solution of differential equation - discontinuity

Typical solution graph using `K=5`.

Example 5

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


Divide throughout by x:



`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`


`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`.


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

Multiplying throughout by x4 gives us `y` as an explicit function of `x`:


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

graph solution of differential equation

Typical solution graph using `K=0.005`.

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