Solve the following equation:
$$
\dfrac {dy}{dx} + \dfrac {y}{x^2}=\dfrac {1}{x^2}
$$
My Attempt
Given:
$$
\dfrac {dy}{dx}+\dfrac {y}{x^2}=\dfrac {1}{x^2}
$$
Comparing above equation with the standard form of first order linear differential equation $dy/dx+P.y=Q$ where $P$ and $Q$ are the functions of $x$ or constants.
Now, using the integrating factor $e^{\int P dx}$:
$$
e^{\int P dx}=e^{\int x^{-2} dx}=e^{-\frac {1}{x}}
$$
Multiplying both sides of the given equation by integrating factor:
$$
e^{-\frac {1}{x}}.\dfrac {dy}{dx}+\dfrac {y.e^{-\frac {1}{x}}}{x^2}=\dfrac {e^{-\frac {1}{x}}}{x^2}
$$
Best Answer
Hint: Write your equation in the form $$\frac{dy}{1-y}=\frac{dx}{x^2}$$ for $$y\ne 1$$ and $$x\ne 0$$