-2.2.1 separable variables y'=\frac{x^2}{y}

[karush]

2000
$\textsf{solve the given differential equation}$
$$y'=\frac{x^2}{y}$$

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]

$\textsf{solve the given differential equation}$
$$y'=\frac{x^2}{y}$$

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]

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*}

 

[Greg]

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]

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

 

[tkhunny]

$\displaystyle \frac{y^2}{2} = \frac{x^3}{3}+c_1$
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]

:-( 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.

 

[HOI]

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]

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]

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]

Exactly. So, why bother to index it? This is my point.

don't know did it last week..