Find the function
y
=
y
(
x
)
\displaystyle {y}={y}{\left({x}\right)}
y
=
y
(
x
)
(for
x
>
0
\displaystyle {x}\gt{0}
x
>
0
) which satisfies the separable differential equation
d
y
d
x
=
4
+
18
x
x
y
2
;
x
>
0
\displaystyle {\frac{{{\left.{d}{y}\right.}}}{{{\left.{d}{x}\right.}}}}={\frac{{{4}+{18}{x}}}{{{x}{y}^{{2}}}}};\ \ \ {x}\gt{0}
d
x
d
y
=
x
y
2
4
+
18
x
;
x
>
0
with the initial condition
y
(
1
)
=
5
\displaystyle {y}{\left({1}\right)}={5}
y
(
1
)
=
5
.
y
=
\displaystyle {y}=
y
=
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