$\frac{dy}{dx}P(x)y=Q(x)y^n$

Observe that, if $n=0$ or $1,$ the Bernoulli equation is linear. For other values of $n,$ the substitution $u=y^{1-n}$ transforms the Bernoulli equation into the linear equation

$\frac{du}{dx}(1-n)P(x)u=(1-n)Q(x)$

Consider the initial value problem

$x{y}^{\text{'}}y=5x{y}^2,y(1)=4$

$($a$)$ This differential equation can be written in the form $(**)$ with

$P(x)=$

$Q(x)=$

$n=$

$$

$,$ and

$$

$($b$)$ The substitution $u=$ will transform it into the linear equation

$\frac{du}{dx}u=$

$($c$)$ Using the substitution in part $($b$),$ we rewrite the initial condition in terms of $x$ and $u$ :

$u(1)=$

$$

$($d$)$ Now solve the linear equation in part $($b$),$ and find the solution that satisfies the initial condition in part $($c$).$

$u(x)=$

$$

$($e$)$ Finally colve for $u$

$y(x)=$

$$

Note: You can earn partial credit on this problem.

$$

$$

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