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

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

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

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

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

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 ?

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