## JEE mains tips and tricks : Linear Differential Equation of First Order

JEE mains tips and tricks : First Order Linear Differential Equation

This might be one of the differential equations of most importance, since many of the applications that you see in physics are modeled by the means of an equations of this type.

## Definition (JEE mains tips and tricks for Linear Equation)

A first order differential equation may be written like so:

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

Where P(x) and Q(x) are real functions, this equation is called linear differential equation.

Note: Take into consideration that a linear differential equation of order n looks something like this:

an (x) y (n) + an-1 (x) y (n-1) + … + a1 (x) y’ + a0 (x) y = f(x)

where the coefficients ai (x) are real functions and an (x) =/= 0. Notice that, when n = 1, we obtain:

a1 (x) y’ + a0 (x) y = f(x)

and if we divide by a1(x):

y’ + [ a0 (x) y / a1 (x) ] = [ f(x) / a1 (x) ]

which is really similar to:

y’ + P(x) . y = Q(x)

where P(x) = [a0 (x) y / a1 (x)] and Q(x) = [f(x) / a1 (x)].

### Theorem

The general solution of the first order differential equation,

y’ + P(x) . y = Q(x) [1]

is given by:

$y=e^{-\int P(x)dx}(\int Q(x)e^{\int P(x)}dx)$

### Demonstration

Rewriting equation [1] as follows:

[P(x) . y – Q(x)] . dx – dy = 0

We can corroborate that $e^{\int P(x)dx}$ it’s an integral factor. Multiplying the equation [1] with this factor we obtain that:

$e^{\int P(x)dx} \frac{dy}{dx} + P(x)y = Q(x)$

in which,

$d\frac{(ye^{\int P(x)dx})}{dx} = Q(x)e^{\int P(x)dx}$

And integrating with respect to x we get that:

$ye^{\int P(x)dx} = \int Q(x)e^{\int P(x)dx}$

which is what we wanted to obtain as a result. Thus, the theorem is demonstrated.

## JEE mains tips and tricks : How to solve exercises?

A linear first order differential equation, written in the standard or canonical form is:

y’ + P(x) . y = Q(x)

If Q(x) = 0, then the equation is homogeneous, on the contrary, if Q(x) =/= 0, then the equation is not homogenerous.

##### Also Read JEE mains tips and tricks : Algebra

It’s almost always possible to resolve analytically a linear differential equation. There is a vast assortment of analytical methods that where devised to solve a first order linear differential equation. The first of those methods is known as the integral factor method, using the following integration or integral factor:

$\mu (x)= e^{\int P(x)dx}$

As you can easily tell, the integration factor depends on the coefficient function of y in the canonical equation, meaning that it depends on P(x).

An alternative while searching the solution of the first order linear differential equation is, taking Q(x) = 0, obtaining in this way the homologous differential equation associated with the canonical equation.

### Integral factor

It is possible to deduce an adequate integral factor, u(x), that facilitates finding the solution of the first order linear differential equation. Let us consider:

y’ + P(x) . y = Q(x)

Firstly, we multiply each term in the equation by u(x) to obtain:

u(x).y’ + u(x).P(x)y = u(x).Q(x)

We then add zero to the left side:

u(x).y’ + u(x).P(x)y + [u’(x).y – u’(x).y] = u(x).Q(x)

Associating in a convenient way:

[u’(x).y + u(x).y’] + [u(x).P(x)y – u’(x).y] = u(x).Q(x)

Identifying and indicating the product derivative and factoring by y the second association:

[u(x).y]’ + [u(x).P(x) – u’(x)]y = u(x).Q(x)

Now, if we consider that:

u(x).P(x) – u’(x) = 0

Then the integral factor that we are searching for is the one that satisfy the above equation.

Substituting, we obtain:

[u(x).y]’ + 0 .y = u(x).Q(x)

[u(x).y]’ = u(x).Q(x)

u(x).y = u(x).Q(x)d(x)

y = [u(x).Q(x)d(x)]/u(x)

The solution to the canonical differential equation is the explicit function given above. The only thing left to do is determining the integral factor u(x).

$\mu (x)p(x)-\mu'(x)=0 \Rightarrow \mu (x)p(x)=\mu'(x) \Rightarrow p(x)= \frac{\mu'(x)}{p(x)}$

$p(x)=\frac{dln\mu(x)}{dx}, \mu(x)>0$

$\int p(x) dx = ln\mu(x) \Rightarrow ln\mu(x)= \int p(x)dx$

### Procedure

To solve a first order linear differential equation, we proceed as follows:

1. We write the equation in it’s canonical form, y’ + P(x)y = Q(x).

2. We identify the coefficient of y, which means, the function P(x) and we determine the integral factor given by e∫ P(x)dx

3. We multiply the equation obtained in step 1by the integral factor calculated in step 2.

e∫ P(x)dx . y’ + e∫ P(x)dx . P(x)y = e∫ P(x)dx . Q(x)

4. We notice that the left side of the equation has the expanded form of the derivative of a product, so we write this derivative in it’s non expanded form:

(e∫ P(x)dx . y)’ = e∫ P(x)dx . Q(x)

(u(x).y)’ = u(x).Q(x)

5. We integrate both members of the equation obtained in step 4:

u(x) . y = u(x).g(x)d(x) + c

6. You clear the function y:

y = [u(x).Q(x)d(x)]/u(x)

Examples

Example 1. Given the equation (x dy/ dx) – 4y = x5ex , find the resolution.

Rewriting the equation we obtain:

(dy/dx) – (4y/x) = x5ex

The integral factor is given by,

u(x) = e-4 ∫ dx/x = e -4 Ln(x) = 1 / x4

knowing this, we can infer that the resolution is given by,

y / x4 = ∫ x e x dx = x e x – e x + c

Meaning that: y = x5ex – x4ex + cx4

is the answer for this particular problem.

##### Also Read JEE mains tips and tricks : Civil Engineering

Example 2. Consider the differential equation,

x2 y dx + [y2 + x2p(x)] dy = 0 [2]

Find a function P(x) so that the differential equation [2] is exact and solve said differential equation.

For the equation [2] to be exact, it must first satisfy:

$\frac{\delta M}{\delta y}= x^{2} = \frac{\delta N}{\delta x} = 2xP(x)+x^{2}P'(x)$

From here, we obtain that the linear differential equation in P and x

x2 = 2xP(x) + x2P’(x) → P’(x) + [2P(x)/x] = 1

which has the following solution:

P(x) = x/3 + c/x2

From there, if we replace c with 0, we obtain that P(x) = x/3, which is the answer.

Example 3.

Check that the differential equation:

y’ + P(x)y = Q(x)y Ln(y)

where P(x) and Q(x) are real functions, it is transformed in a linear differential equation when u = Ln(y).

Then, all that is left to do is:

u = Ln(y) → u’ = y’/y → y’ = u’y = u’eu

Substituting in the given equation:

y’ + P(x)y = Q(x)y Ln(y)

u’eu + P(x)eu = Q(x)ueu

u’ + P(x) = Q(x)u

u’ – Q(x)u = – P(x)

which is a linear differential equation.