Help with calc please?

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Help with calc please?

Use partial fractions to solve the separable equation dy/dx=-2y/x^2-1

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Answer 1 · 2018-07-04T14:35:59

I’ll interpret the equation as $\frac{d y}{d x} = - \frac{2 y}{x^{2}} - 1$ $dy/dx = -(2y)/x^2 -1$

Explanation:

Let’s start with a small rearrangement.

$\frac{d y}{d x} + \frac{2 y}{x^{2}} = - 1$ $dy/dx + (2y)/x^2 =-1$

We now can see the equation is in the form $\frac{d y}{d x} + f (x) y = g (x)$ $dy/dx + f(x)y = g(x)$ , a linear differential equation. The first step here is to find the integrating factor which will always be given by $e^{\int f (x) d x}$ $e^(intf(x) dx)$ .

It follows that $I = e^{\int \frac{2}{x^{2}} d x} = e^{- \frac{2}{x}}$ $I= e^(int 2/x^2dx) = e^(-2/x)$

Multiply both sides of the equation by $e^{- \frac{2}{x}}$ $e^(-2/x)$

$\frac{d y}{d x} e^{- \frac{2}{x}} + e^{- \frac{2}{x}} \frac{2 y}{x^{2}} = - e^{- \frac{2}{x}}$ $dy/dxe^(-2/x) + e^(-2/x)(2y)/x^2 = -e^(-2/x)$

$(ye^(-2/x))’ = e^(-2/x)$

$ye^(-2/x) = int e^(-2/x) dx + C$

However, $int e^(-2/x) dx$ can’t be integrated in terms of elementary functions so it is completely allowable for us to leave it as this in the final answer.

$y = (int e^(-2/x)dx+ C)/e^(-2/x)$

Hopefully this helps!

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Help with calc please?

Use partial fractions to solve the separable equation dy/dx=-2y/x^2-1

1 Answer

Explanation:

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