Answer
a) $α(x)$ = $x$
b) $(xy)'$ = $x^{4}$
c) $y$ = $\frac{1}{5}x^{4}+\frac{C}{x}$
d) $y$ = $\frac{1}{5}x^{4}-\frac{1}{5x}$
Work Step by Step
a)
the equation is of the from
$y'+A(x)y$ = $B(x)$
$A(x)$ = $x^{-1}$
$B(x)$ = $x^{3}$
by theorem 1, $α(x)$ is defined by
$α(x)$ = $e^{\int{A(x)}dx}$ = $e^{\ln{x}}$ = $x$
b)
multiplied by $α(x)$, the equation is
$xy'+y$ = $x^{4}$
$(xy)'$ = $x^{4}$
c)
$(xy)'$ = $x^{4}$
$(xy)$ = $\frac{1}{5}x^{5}+C$
$y$ = $\frac{1}{5}x^{4}+\frac{C}{x}$
d)
if $y(1)$ = $0$
$0$ = $\frac{1}{5}+C$
$C$ = $-\frac{1}{5}$
so
$y$ = $\frac{1}{5}x^{4}-\frac{1}{5x}$
