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# 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

(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 x3 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.

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

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:

ye3x = 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

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 x2:

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

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

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 x4 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.