eBernoulliDFQ

Example Solve the following initial value problem.

$\displaystyle xy^{\prime} - y = -y^{2}, y(1) = 1$

Write this in the standard form.

$\displaystyle y^{\prime} - \frac{1}{x} y = -\frac{1}{x}y^{2}$

This is a Bernoulli's equation. Then we multiply $y^{-2}$ to both sides of the equation and simplify to get

$\displaystyle y^{-2}y^{\prime} - \frac{1}{x} y^{1-2} = -\frac{1}{x}$

Now we put $u = y^{1-2} = y^{-1}$ . Then $u^{\prime} = -y^{-2}y^{\prime}$ and

$\displaystyle - u^{\prime} - \frac{1}{x} u = - \frac{1}{x}$

This differential equation is linear in $u$

$u$

. Rewrite this in the standard form in $u$

$u$

と

$\displaystyle u^{\prime} + \frac{1}{x} u = \frac{1}{x}$

Now find the integrating factor $\mu$ . Then $\mu = \exp(\int (1/x) dx) = \exp(\log{x}) = x$ . We multiply $\mu$ to both sides of the equation. Then the left-side is a derivative of the product of $\mu$ times the dependent variable $x$

$x$

. Thus we have

$\displaystyle ( x u)^{\prime} = 1$

We integrate the above equation with respect to $x$

$x$

$\displaystyle xu = \int 1 dx = x + c$

$\displaystyle u = \frac{x + c}{x}$

Since $u = y^{-1}$ , we have

$\displaystyle y^{-1} = \frac{x + c}{x}$

Here we apply the initial condition $y(1) = 1$

$y(1) = 1$

to obtain

$\displaystyle y^{-1} = \frac{x + 1}{x}$

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