$
x=t^2$

$\frac{dx}{dy}=\frac{dx}{dt}\cdot\frac{dt}{dy}$

$\frac{dx}{dy}=2t\cdot\frac{dt}{dy}$

$\frac{dy}{dx}=\frac{1}{2t}\cdot\frac{dy}{dt}$

NOTE I am solving a Second Order Differential Equation via substitution, thus I shall differentiate again - using the product rule - below I have differentiated the two multiples of $\frac{dy}{dx}$ separately and then added them together in the end to form $\frac{d^2y}{dx^2}$

$\frac{d}{dx}(\frac{1}{2t})=-\frac{1}{2t^2}\cdot\frac{dt}{dx}$

$\frac{d}{dx}\frac{dy}{dt}=\frac{d^2y}{dt^2}\cdot\f rac{dt}{dx}$

$\frac{d^2y}{dx^2}=\frac{1}{2t}\cdot\frac{d^2y}{dt^ 2}\cdot\frac{dt}{dx}-\frac{1}{2t^2}\cdot\frac{dt}{dx}\cdot\frac{dy}{dt}$

Am I correct?
If not please do explain where and how I am wrong.
If I am correct, please reassure me on why I have done this, I don't fully understand the whole concept.

2. Originally Posted by iPod
$
x=t^2$

$\frac{dx}{dy}=\frac{dx}{dt}\cdot\frac{dt}{dy}$

$\frac{dx}{dy}=2t\cdot\frac{dt}{dy}$

$\frac{dy}{dx}=\frac{1}{2t}\cdot\frac{dy}{dt}$

NOTE I am solving a Second Order Differential Equation via substitution, thus I shall differentiate again - using the product rule - below I have differentiated the two multiples of $\frac{dy}{dx}$ separately and then added them together in the end to form $\frac{d^2y}{dx^2}$

$\frac{d}{dx}(\frac{1}{2t})=-\frac{1}{2t^2}\cdot\frac{dt}{dx}$

$\frac{d}{dx}\frac{dy}{dt}=\frac{d^2y}{dt^2}\cdot\f rac{dt}{dx}$

$\frac{d^2y}{dx^2}=\frac{1}{2t}\cdot\frac{d^2y}{dt^ 2}\cdot\frac{dt}{dx}-\frac{1}{2t^2}\cdot\frac{dt}{dx}\cdot\frac{dy}{dt}$

Am I correct?
If not please do explain where and how I am wrong.
If I am correct, please reassure me on why I have done this, I don't fully understand the whole concept.

Dear iPod,

You are correct. I don't know why you have done this since the differential equation you are solving is not stated here. But if you give out the complete solution to the differential equation (or what you have done) I may be able to help you.

3. I apologize, the question says;

Show that the substitution $x=t^2$ transforms the differential equation

$4x\frac{d^2y}{dx^2}+2(1-2\sqrt{x})\frac{dy}{dx}+y=3\sqrt{x}$

into

$\frac{d^2y}{dt^2}-2\frac{dy}{dt}+y=3t$
__________________________________________________ _______

So I presume if I am correct, I just plug in my values of $x , \frac{dy}{dx} , \frac{d^2y}{dx^2}$ in terms of $t$ into my original equation, yeah?

4. Originally Posted by iPod
I apologize, the question says;

Show that the substitution $x=t^2$ transforms the differential equation

$4x\frac{d^2y}{dx^2}+2(1-2\sqrt{x})\frac{dy}{dx}+y=3\sqrt{x}$

into

$\frac{d^2y}{dt^2}-2\frac{dy}{dt}+y=3t$

__________________________________________________ _______

So I presume if I am correct, I just plug in my values of $x , \frac{dy}{dx} , \frac{d^2y}{dx^2}$ in terms of $t$ into my original equation, yeah?

Dear iPod,

Yes you can do that. If you have any problem please don't be hesitant to discuss.

5. Cool, I have used it and ended up with the correct answer, thank you.

However, could you just talk me through about differentiating $t$ in respect to $x$ so that I have a clearer idea of what I am doing, because it all seems pretty ambiguous. I'm sure it is related to the Chain Rule and Implicit Differentiation.

Thank you

6. Originally Posted by iPod
Cool, I have used it and ended up with the correct answer, thank you.

However, could you just talk me through about differentiating $t$ in respect to $x$ so that I have a clearer idea of what I am doing, because it all seems pretty ambiguous. I'm sure it is related to the Chain Rule and Implicit Differentiation.

Thank you
Dear iPod,

$x=t^2$

$\frac{dx}{dy}=2t\frac{dt}{dy}\Rightarrow{\frac{dy} {dx}=\frac{1}{2t}\frac{dy}{dt}}$

$\frac{d^{2}y}{dx^2}=-\frac{1}{2t^2}\frac{dt}{dx}\frac{dy}{dt}+\frac{1}{ 2t}\frac{d^{2}y}{dt^2}\frac{dt}{dx}=-\frac{1}{2t^2}\frac{1}{2t}\frac{dy}{dt}+\frac{1}{2 t}\frac{d^{2}y}{dt^2}\frac{1}{2t}=\frac{1}{4t^2}\f rac{d^{2}y}{dt^2}-\frac{1}{4t^3}\frac{dy}{dt}$

Do you understand this? The rules used here are the chain rule and the product rule of differentiation. If you have any problems don't hesitate to reply.

7. Ah yes, that makes it much clearer for me, I've actually understood what is going on. I know what im doing now :P

Thank you very much