I Separation of variables and the chain rule

[Martyn Arthur]

TL;DR

Application of chain rule

Hi; given the equation ydy/dx=x^2 how is the chain rule applied to result in ydy =x^2dx?
Thanks

 

[DeBangis21]

What I know is that is a 1st order ODE can be solved by 'Seperating the Variables' method. So, #ydy=x^2dx# just take the integral of both side.

Is that what you meant in the OP?

 

[Martyn Arthur]

sorry no; I don't understand the actual application of the chain rule to produce the result

 

[DeBangis21]

Ok, I believe the experts are coming along to help out.

 

[PeroK]

TL;DR Summary: Application of chain rule

Hi; given the equation ydy/dx=x^2 how is the chain rule applied to result in ydy =x^2dx?
Thanks

That's not the chain rule. That's the defining property of the differentials $dx$ and $dy$.

 

[PeroK]

You could use the chain rule as follows:
$$\frac d {dx}\big (y^2\big ) = 2y \frac{dy}{dx} = 2x^2$$$$\implies y^2 = \int 2x^2 dx$$

 

[pasmith]

\begin{split}<br /> y\frac{dy}{dx} &amp;= x^2 \\<br /> \int y\frac{dy}{dx} \,dx &amp;= \int x^2 \,dx \end{split} The left hand side can be integrated by substitution, also known as the "!inverse chain rule" since <br /> \int_a^b g&#039;(f(x))f&#039;(x)\,dx = \int_a^b (g \circ f)&#039;(x) \,dx = g(f(b)) - g(f(a)) = \int_{f(a)}^{f(b)} g&#039;(y)\,dy by the chain rule and the fundamental theorem.

 

[Mark44]

Hi; given the equation ydy/dx=x^2 how is the chain rule applied to result in ydy =x^2dx?

No chain rule at all -- what they did was to multiply both sides of the equation by dx.

 

[Martyn Arthur]

ah; thanks