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[SOLVED] 2.2.1 separable equations

karush

Well-known member
Jan 31, 2012
2,928
$\textsf{solve the given differential equation}$
$$y'=\frac{x^2}{y}$$

ok this is a new section on separable equations
so i barely know anything
but wanted to post the first problem
hoping to understand what the book said.
thanks ahead...
 

Sudharaka

Well-known member
MHB Math Helper
Feb 5, 2012
1,621
$\textsf{solve the given differential equation}$
$$y'=\frac{x^2}{y}$$

ok this is a new section on separable equations
so i barely know anything
but wanted to post the first problem
hoping to understand what the book said.
thanks ahead...
Hi,

This is a separable differential equation; and more information about separable differential equations are explained here,

Differential Equations - Separable Equations

To solve this you can write it as,

$$y\frac{dy}{dx}=x^2$$

$$\Rightarrow \int y dy = \int x^2 dx$$

Hope you can continue from here.
 

karush

Well-known member
Jan 31, 2012
2,928
ok I presume this but $c_1$ and $c_2$ ???
\begin{align*}
\int y \, dy &= \int x^2 \, dx \\
\frac{y^2}{2} & = \frac{x^3}{3}
\end{align*}
 
Last edited:

Greg

Perseverance
Staff member
Feb 5, 2013
1,404
ok I presume this but $c_1$ and $c_2$ ???
\begin{align*}
\int y \, dy &= \int x^2 \, dx \\
\frac{y^2}{2}+c_1 & = \frac{x^3}{3}+c_2
\end{align*}
Correct. Now combine the constants $c_1$ and $c_2$ into a single constant on the RHS and solve for $y$.
 

karush

Well-known member
Jan 31, 2012
2,928
$\displaystyle \frac{y^2}{2} = \frac{x^3}{3}+c_1$
cross mulitply
$\displaystyle 3y^2-2x^3=c_1$
 

tkhunny

Well-known member
MHB Math Helper
Jan 27, 2012
267
$\displaystyle \frac{y^2}{2} = \frac{x^3}{3}+c_1$
cross mulitply
$\displaystyle 3y^2-2x^3=c_1$
:-( We can never be friends.

If you are going to go to the trouble to index your constants, you probably should change the index when the nature of the constant changes.
 

karush

Well-known member
Jan 31, 2012
2,928
:-( We can never be friends.

If you are going to go to the trouble to index your constants, you probably should change the index when the nature of the constant changes.
the text index's the constants no matter what. 😰
 

Country Boy

Well-known member
MHB Math Helper
Jan 30, 2018
756
What text is that?

If [tex]\frac{y^2}{2}= \frac{x^3}{3}+ c_1[/tex] then multiplying both sides by 6 gives

either

[tex]3y^2= 2x^3+ 6c_1[/tex] or [tex]3y^2= 2x^3+ c_2[/tex].
 

karush

Well-known member
Jan 31, 2012
2,928
What text is that?

If [tex]\frac{y^2}{2}= \frac{x^3}{3}+ c_1[/tex] then multiplying both sides by 6 gives

either

[tex]3y^2= 2x^3+ 6c_1[/tex] or [tex]3y^2= 2x^3+ c_2[/tex].
the book gave $3{y}^{2}-2x^3=c$ so there was no need to find c
 

tkhunny

Well-known member
MHB Math Helper
Jan 27, 2012
267
the book gave $3{y}^{2}-2x^3=c$ so there was no need to find c
Exactly. So, why bother to index it? This is my point.
 

karush

Well-known member
Jan 31, 2012
2,928