# Find an integrating factor for the linear differential equation and hence solve the equation: dy/dx-2y/x=x^2*sinxNeed answers by 15th May! thanks!

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Notice that the integrating factor is the coefficient of y such that:

`mu` (x) `= e^(int (-2/x)dx) = e^(-2ln x)`

`mu (x) = e^(ln (x^(-2)))`

`mu (x) = x^(-2) =gt mu (x) = 1/x^2`

You need to solve the differential equation, hence, you need to multiply the equation by the integrating factor `1/x^2` such that:

`(1/x^2)*y' - (2/x^3)y = sin x`

Notice that the left side is the expansion of `((1/x^2)*y)'` using the product rule such that:

`((1/x^2)*y)' = sin x`

Integrating both sides yields:

`int ((1/x^2)*y)' = int sin xdx`

`(1/x^2)*y = - cos x + c`

You need to multiply by `x^2` such that:

`y = -x^2*cos x + x^2*c`

**Hence, evaluating the integrating factor yields `mu (x) = 1/x^2` and evaluating the solution to differential equation yields `y = -x^2*cos x + x^2*c` .**