Question

Writ the Integrating factor of differential equation

xdy/dx -2y= x^2 + sin(1/x^2)

Answer 1

Answer:

$\frac{1}{x^2}$

Step-by-step explanation:

The given differential equation

$x\frac{dy}{dx}-2y=x^2+\sin\frac{1}{x^2}$

or, $\frac{dy}{dx}-\frac{2}{x}y=x+\frac{1}{x}\sin\frac{1}{x^2}$

We know that for differential equation of the form

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

The integrating factor is given by

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

Therefore, for the given differential equation, the integrating factor is

$I=e^{\int (-2/x)dx}$

$\implies I=e^{-2\int (1/x)dx }$

$\implies I=e^{-2\ln x}$

$\implies I=e^{\ln x^{-2}}$    

$\implies I=x^{-2}$           (∵ $e^{\ln t}=t$)

$\implies I=\frac{1}{x^2}$

Hope this answer is helpful.

Answer 2

Answer:

Therefore the integrating factor is $\frac{1}{x^{2}}$

Step-by-step explanation:

In this question,

We have been given the differential equation

$x\frac{dy}{dx}\ -\ 2y\ =\ x^{2}\ +\ sin\frac{(1)}{(x^{2})}$

We need to find the integrating factor of this linear differentail equation

Firstly we need to make this a linear differential equation,

Dividing whole equation by x we get,

$\frac{dy}{dx}\ -\ \frac{2y}{x}\ =\ x\ +\ \frac{sin\frac{(1)}{(x^{2})}}{x}$

We know the basic form of differential equation

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

Integrating factor is given by

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

Here p(x) = $\frac{-2}{x}$

Therefore, the integrating factor is

I = $e^{\int (-2/x)dx}$

$\implies I\ =\ e^{-2\int (1/x)dx }$

$\implies I\ =\ e^{-2\ln x}$

$\implies I\ =\ e^{\ln x^{-2}}$    

$\implies I\ =\ x^{-2}$           (∵ $e^{\ln t}\ =\ t$)

$\implies I\ =\ \frac{1}{x^2}$

Therefore the integrating factor is $\frac{1}{x^{2}}$