Calculate the \[\mu\] : Multiply the differential equation with integrating factor on both sides in such a way; \[\mu \frac{dy}{dx} + \mu P(x)y = \mu Q(x)\] In this way, on the left-hand side, we obtain a particular differential form.

The first step is to calculate and find out the integrating factor, let us say μ. Now we have to make the whole differential function in a product form with the integrating factor on both the LHS and RHS that is for example-

\[\mu \frac{di}{dz} + \mu R(x)y = \mu T(x)\].

This is the process of integrating factor method and the consecutive steps to use the integrating factor method formula.

The general form of representing the linear differential equations is-

y’ + G y=K

or

dy/dz + G(z) y =K(z)

Where y’ or dy/dz is the first derivative. Also, the functions G and K are the functions of z only.

There are mainly two methods which are utilized to solve the linear first-order differential equations: In mathematical calculus, two main methods which are used to equate the linear first-order differential equation are-

Separable Method

Integrating Factor Method

Now we will see and understand how we can apply these two methods and calculate the integrating factor accurately.

Integrating Factor Method technique of solving applied to solve the first and second-order differential equations.

We now know what integrating factor means, and it is to be calculated to solve the inexact differential equation. This method is generally used in the case of linear differential equations and sometimes can be used for higher differential calculus as well.

When the differential equation is in the form of:

\[\frac{(dy)}{(dx)} + P(x)(y) = Q(x)\]

then the integrating factor can be defined as:

First order-

We need to Compare the given equation with the differential equation form and find the value of G(z).

Now we need to start Calculating the integrating factor μ by multiplying the differential equation with the integrating factor.

By multiplying the differential equation with integrating factor on both LHS and RHS in a prescribed way, we get-

\[\mu \frac{dy}{dz} + \mu P(z)(y) = \mu K(z)\]

By calculating like this, on the LHS, we get a particular differential form. Which is\[\frac{d}{dz} (\mu y) = \mu K(z)\].

At last, we should integrate the differential equation we have calculated using the basic formulas of integration and obtain the necessary result to the equation which is already given:

μ y = \[\int \mu G(z)dz\] + Constant term.

The integrating factor method can be easily used to solve the second-order differential equation.

Let the provided differential equation be written as:

y” + G(z) y’ = K(z);

The second-order equation given above can be easily solved by the integrating factor method.

Substitute y’ = v; so that the equation becomes similar to the first-order equation as shown:

v’ + G(z) v = K(z)

Now, this equation is required to be arranged and solved by integrating factor method as elaborated in the subheading placed above for first-order equations, and we finally get the equation: μ v=∫μG(z)dz + Constant term

Find the value of v from the resulted equation.

Since v = y’, hence to find the value of y, integrate the equation. In this way, we get the required solution.

Therefore, to find out the exact value of y, we need to integrate the resulted equation again. By following the above-mentioned methods, we can get the necessary result for the given equation.

Solve the differential equation using the integrating factor method formula: \[(\frac{dz}{dx}) - (\frac{3z}{x+1}) = (x+1)^{4}\]

Solution:

Given: \[(\frac{dz}{dx}) - (\frac{3z}{x+1}) = (x+1)^{4}\]

First, we need to find the integrating factor by using the concepts of

\[\mu = e\int g(x)dx\]

\[\mu = e\int (\frac{-3}{x+1})dx\]

\[\int (\frac{-3}{x+1})dx = -3 log(x+1)^{-3}\]

therefore, we get

\[\mu = e log(x+1)^{-3}\]

\[\mu = \frac{1}{(x+1)^{3}}\]

Now, we need to multiply the integrating factor on both the LHS and RHS of the provided differential equation:

\[[\frac{1}{(x+1)^{3}}] [\frac{dz}{dx}] - [\frac{3z}{(x+1)^{4}}]\] = (x+1)

Integrating both the sides, we get:

\[[\frac{z}{(x+1)^{3}}] = [(\frac{1}{2})x^{2} + x + \text{constant term}]\]

Therefore, the general solution of the provided differential equation is

\[z = [(x+1)^{3}] = [(\frac{1}{2})x^{2} + x + \text{constant term}]\].

FAQ (Frequently Asked Questions)

1. When can we use the Core Differential Equation Integrating Factor Method?

Ans. The integrating factor is generally used as a multiplying factor to solve complicated inexact differential equations of the first order. Using this integrating factor as a multiplier, the function can be converted to the exact first-order differential equation easily, and by doing that, the equation can be solved very easily and smoothly. They are generally used for first-order differential equations but can also be used for higher-order calculus operations. We use the integrating factors when we are required to convert the differential equations into suitable forms for calculus calculations. Hence in this way, the procedure of solving the equation can be done.

2. How can we Conclude that a Function is Linear or not?

Ans-The way to tell if an equation is linear is by using y = gx + c. where g and b can stand to any constant as long it stays the same number for the existing equation. If the equation has a structure like the mentioned above, then the equation is always linear. If an equation also has a structure similar to gx = y, then the equation is linear. Once we know the equation is linear, we can easily find the integrating factor and solve the provided equation. This method can be used for thermodynamic equations as well, so it can be broadened to the subject of physics too.