Question

Find the integrating factor of the differential equation x dy/dx-2y=2x^2

Answer 1

The integrating factor is $\dfrac{1}{x^2}$

Step-by-step explanation:

Since we know that;

•   $\dfrac{dy}{dx} +Py = Q$  

then;  

• I. F. = $e^\int^p^d^x$

Differential equation:

• $x \dfrac{dy}{dx}-2y=2x^2$

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

P = $\dfrac{-2}{x}$, Q = 2x

Now, I.F. =  $e^\int^p^d^x$

               =$e^{\int\dfrac{-2}{x}dx}$

               = $e^{-2logx}$

               = $e^{log\left(x\right )^-^2}$

               = $x^-^2$

Therefore,  I.F. = $\dfrac{1}{x^2}$

Answer 2

The integrating factor is $\dfrac{1}{x^2}$

Step-by-step explanation:

Given that,

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

This is linear differential equation, y' + Py = Q

then Integrating factor, $IF=e^{\int Pdx}$

First we change the differential equation,

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

  y ' +py = Q

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

where, $P=-\dfrac{2}{x},Q=2x$

Integrating factor, $IF=e^{\int -\dfrac{2}{x}dx}$

                              $IF=e^{\ln x^{-2}}$

                              $IF=x^{-2}=\dfrac{1}{x^2}$

Hence, the integrating factor is $\dfrac{1}{x^2}$

#Learn more:

Integrating factor

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