Find an implicit solution to $y ^ { prime } ( x ) = frac { x ^ { 3 } } { 2 y sqrt { 1 + y ^ { 2 } } }$












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 ?






integration







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$endgroup$








  • 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 at 0:39










  • $begingroup$
    SO its done even with the + C ?
    $endgroup$
    – Tariro Manyika
    Jan 14 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 at 0:41










  • $begingroup$
    You cannot find an explicit formula without an initial condition.
    $endgroup$
    – Tom Himler
    Jan 14 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)
    $endgroup$
    – Tariro Manyika
    Jan 14 at 0:51
















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 ?






integration







share|cite|improve this question









$endgroup$








  • 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 at 0:39










  • $begingroup$
    SO its done even with the + C ?
    $endgroup$
    – Tariro Manyika
    Jan 14 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 at 0:41










  • $begingroup$
    You cannot find an explicit formula without an initial condition.
    $endgroup$
    – Tom Himler
    Jan 14 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)
    $endgroup$
    – Tariro Manyika
    Jan 14 at 0:51














5












5








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 ?






integration







share|cite|improve this question









$endgroup$




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



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



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



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



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 ?






integration




integration






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Jan 14 at 0:35









Tariro ManyikaTariro Manyika

619




619








  • 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 at 0:39










  • $begingroup$
    SO its done even with the + C ?
    $endgroup$
    – Tariro Manyika
    Jan 14 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 at 0:41










  • $begingroup$
    You cannot find an explicit formula without an initial condition.
    $endgroup$
    – Tom Himler
    Jan 14 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)
    $endgroup$
    – Tariro Manyika
    Jan 14 at 0:51














  • 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 at 0:39










  • $begingroup$
    SO its done even with the + C ?
    $endgroup$
    – Tariro Manyika
    Jan 14 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 at 0:41










  • $begingroup$
    You cannot find an explicit formula without an initial condition.
    $endgroup$
    – Tom Himler
    Jan 14 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)
    $endgroup$
    – Tariro Manyika
    Jan 14 at 0:51








1




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 at 0:39




$begingroup$
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 at 0:39












$begingroup$
SO its done even with the + C ?
$endgroup$
– Tariro Manyika
Jan 14 at 0:40




$begingroup$
SO its done even with the + C ?
$endgroup$
– Tariro Manyika
Jan 14 at 0:40




1




1




$begingroup$
In order to find the value of $c$, you need to be provided an initial condition.
$endgroup$
– D.B.
Jan 14 at 0:41




$begingroup$
In order to find the value of $c$, you need to be provided an initial condition.
$endgroup$
– D.B.
Jan 14 at 0:41












$begingroup$
You cannot find an explicit formula without an initial condition.
$endgroup$
– Tom Himler
Jan 14 at 0:41




$begingroup$
You cannot find an explicit formula without an initial condition.
$endgroup$
– Tom Himler
Jan 14 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)
$endgroup$
– Tariro Manyika
Jan 14 at 0:51




$begingroup$
If I may ask , is my working correct ? (P.S. took me ages to put on latex inorder to ask this question)
$endgroup$
– Tariro Manyika
Jan 14 at 0:51










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