5
$\begingroup$

$y ^ { \prime } ( x ) = \frac { x ^ { 3 } } { 2 y \sqrt { 1 + y ^ { 2 } } }$

$\frac { d y } { d x } = \frac { x ^ { 3 } } { 2 y \sqrt { 1 + y ^ { 2 } } }$

$\frac { d x } { d y } = \frac { 2 y \sqrt { 1 + y ^ { 2 } } } { x ^ { 3 } }$

$\int x ^ { 3 } d x = 2 \int y \left( 1 + y ^ { 2 } \right) ^ { 1 / 2 } d y$

R.H.S.

$\int y \sqrt { 1 + y ^ { 2 } } d y$

let $u = 1 + y ^ { 2 }$

$ d u = 2 y d y $

$ \frac { d u } { 2 } = y d y $

$\int y \sqrt { 1 + y ^ { 2 } } d y$

$\int \sqrt { 1 + y ^ { 2 } } y d y$

$\int \frac { \sqrt { u } } { 2 } d u$

$\frac { 1 } { 2 } \int \sqrt { u } d u$

$= \frac { 1 } { 2 } \left[ \frac { 2 u ^ { 3 / 2 } } { 3 } \right]$

$= \frac { u ^ { 3 / 2 } } { 3 } = \frac { 1 } { 3 } \left( 1 + y ^ { 2 } \right) ^ { 3 / 2 } \ldots 1$

$\int x ^ { 3 } d y = 2 \int y \left( 1 + y ^ { 2 } \right) ^ { 1 / 2 } d y$

$\frac { x ^ { 4 } } { 4 } + c = 2 \left[ 1 / 3 \left( 1 + y ^ { 2 } \right) ^ { 3 / 2 } \right]$

$\frac { x ^ { 4 } } { 4 } + c = \frac { 2 } { 3 } \left( 1 + y ^ { 2 } \right) ^ { 3 / 2 }$

cant finish it , or maybe my working is wrong ?

$\endgroup$
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  • 1
    $\begingroup$ It looks like you already have an implicit solution. Remember that implicit means you do not need to isolate the variable $y$. $\endgroup$
    – D.B.
    Jan 14, 2019 at 0:39
  • $\begingroup$ SO its done even with the + C ? $\endgroup$ Jan 14, 2019 at 0:40
  • 1
    $\begingroup$ In order to find the value of $c$, you need to be provided an initial condition. $\endgroup$
    – D.B.
    Jan 14, 2019 at 0:41
  • $\begingroup$ You cannot find an explicit formula without an initial condition. $\endgroup$
    – Tom Himler
    Jan 14, 2019 at 0:41
  • $\begingroup$ If I may ask , is my working correct ? (P.S. took me ages to put on latex inorder to ask this question) $\endgroup$ Jan 14, 2019 at 0:51

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