# [SOLVED]2.2.1 separable equations

#### karush

##### Well-known member
$\textsf{solve the given differential equation}$
$$y'=\frac{x^2}{y}$$

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.

#### Sudharaka

##### Well-known member
MHB Math Helper
$\textsf{solve the given differential equation}$
$$y'=\frac{x^2}{y}$$

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.
Hi,

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
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*}

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#### Greg

##### Perseverance
Staff member
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
$\displaystyle \frac{y^2}{2} = \frac{x^3}{3}+c_1$
cross mulitply
$\displaystyle 3y^2-2x^3=c_1$

#### tkhunny

##### Well-known member
MHB Math Helper
$\displaystyle \frac{y^2}{2} = \frac{x^3}{3}+c_1$
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
:-( 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
What text is that?

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

either

$$3y^2= 2x^3+ 6c_1$$ or $$3y^2= 2x^3+ c_2$$.

#### karush

##### Well-known member
What text is that?

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

either

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

#### tkhunny

##### Well-known member
MHB Math Helper
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
Exactly. So, why bother to index it? This is my point.
don't know did it last week..