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!
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` .