# 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

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

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

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

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

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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