What is the general solution of the differential equation? : $\frac{d y}{d x} = \frac{1}{x^{2}}$ $dy/dx = 1/x^2$

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Answer 1 · 2018-02-07T18:03:56

$y = - \frac{1}{x} + C$ $y =-1/x + C$

We have:

$\frac{d y}{d x} = \frac{1}{x^{2}}$ $dy/dx = 1/x^2$

This is a First Order Separable Ordinary Differential Equation. We can "separate the variables" to get:

$int \ dy = int \ 1/x^2 \ dx$

Both integrals are standard calculus results , so integrating we get:

$y = x^(-1)/(-1) + C$
$\ \ = -1/x + C$

Which is the General Solution

What is the general solution of the differential equation? : dy/dx = 1/x^2 dydx=1x2 dy/dx = 1/x^2