Question

how to find integrating factor of dy/dx- 2y/x= x^2​

Answer 1

Answer:

The integrating factor for the given expression will be:

$I.F=\dfrac{1}{x^2}$

Step-by-step explanation:

The given function is:

$\dfrac{dy}{dx}-\dfrac{2y}{x}=x^2$

We can write as:

$\dfrac{dy}{dx}=y'$

Hence the above expression will be written as:

$y'-\dfrac{2y}{x}=x^2$

The above function is in the form of first order linear differential equation.

Hence the general form of first order linear differential equation is:

$y'+P(x)y=Q(x)$

So, we can write as:

$P(x)=\dfrac{-2}{x}\\Q(x)=x^2$

Therefore integrating factor (I.F ) will be:

$I.F=e^{\int P(x)\ dx}\\I.F=e^{\frac{-2}{x}\ dx}$

As we know the formula of:

$\displaystyle\int \dfrac{1}{x}dx=logx$

Hence the above function will be:

$I.F=e^{-2logx}$

Also we know the formula of logarithm that has been shown below:

$a\ logb=logb^a$

Therefore:

$I.F=e^{log(x^{-2})}$

And,

$e^{log}=1$

So,

$I.F=x^{-2}\\I.F=\dfrac{1}{x^2}$