Solve the following ordinary differential equation by the integrating factor met

Question

Solve the following ordinary differential equation by the integrating factor method. dy/dx + 2y/x = x-2 and y(1) = 2 Observe that this is a first order linear differential equation because it can be written in the form y' + f(x) y = g(x) When this happens one should calculate the integration factor f(x) dx Phi(x) = e What is the integrating factor Phi(x) in this case? Phi(x) = Enter just an expression for Phi(x). What did you calculate for the following antiderivative? g(x) Phi(x) dx = Enter just an expression. What is the solution to the initial value problem? y(x) = Enter just an expression for y(x), do not enter an equation.

Explanation / Answer

1)so f(x) = 2/x
so integral of 2/x = 2 ln x = ln x^2
so integrating factor = e^ ln x^2 = x^2
2)
g(x) = x^-2
so g(x)(x)= x^2 * x^-2 = 1

so integral of g(x)(x) = x+C

3) so we know

x^2 y = x + C

so y = 1/x + C/x^2

use initial condition y(1)=2

1 + C = 2 so C= 1

y= 1/x + 1/x^2

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