Thread: Verifying by substitution.

1. Verifying by substitution.

Here's the question:

Verify by substitution that $\displaystyle y(t)=4sin(t^2)-3cos(t^2)$is a solution to $\displaystyle t \frac{d^2y}{dt^2}-\frac{dy}{dt}+4t^3 y=0, t>0.$

I decided to do this question by differentiating y twice (to get$\displaystyle \frac{d^2y}{dt^2}$ and $\displaystyle \frac{dy}{dt}$) and subsituting these expressions in. Then I would rearrange the equation to make y the subject and prove that what I get is $\displaystyle 4sint^2-3cost^2$.

$\displaystyle y=4sint^2-3cost^2$

$\displaystyle \frac{dy}{dt}=8tcos(t^2)+6tsin(t^2)$

$\displaystyle \frac{d^2y}{dt^2}=-16t^2 sin(t^2)+12t^2 cos(t^2)$

After substituting these expressions in I end up with:

$\displaystyle y=4sin(t^2)-\frac{3}{2t^2}sin(t^2)-3cos(t^2)+\frac{2}{t^2}cos (t^2)$.

I just wanted to check if my differentiation was correct. I have double and triple checked my substitution and that's fine.

It just occurred to be that I should have put this in the calculus section. Sorry!

2. Hello,

Hmmm I don't quite understand what you want to do!

Your expressions for $\displaystyle \frac{dy}{dt}$ and $\displaystyle \frac{d^2y}{dt^2}$ are correct.

Now, substitute them in $\displaystyle t \frac{d^2y}{dt^2}-\frac{dy}{dt}+4t^3 y$ and prove it equals 0.

3. That's what I thought as well. Here's how my substitution goes:

$\displaystyle t(-16t^2 sin(t^2)+12t^2 cos (t^2))-(8tcos(t^2)+6tsin(t^2))+4t^3 (4sin (t^2)-3cos(t^2))$

$\displaystyle 4t^3(-4sin(t^2)+3cos(t^2))-2t(4cos(t^2)+3sin(t^2))+4t^3(4sin(t^2)-3cos(t^2))$

$\displaystyle 4t^3(-4sin(t^2)+3cos(t^2)+4sin(t^2)-3cos(t^2))-2t(4cos(t^2)+3sin(t^2))$

$\displaystyle =-2t(4cos(t^2)+3sin(t^2))=-\frac{dy}{dt}$

I did it a different way to how I did before and I now have this answer. I can't see what i'm doing wrong or if it's a typo on the homework sheet.

4. Originally Posted by Showcase_22
That's what I thought as well. Here's how my substitution goes:

$\displaystyle t(-16t^2 sin(t^2)+12t^2 cos (t^2))-(8tcos(t^2)+6tsin(t^2))+4t^3 (4sin (t^2)-3cos(t^2))$

$\displaystyle 4t^3(-4sin(t^2)+3cos(t^2))-2t(4cos(t^2)+3sin(t^2))+4t^3(4sin(t^2)-3cos(t^2))$

$\displaystyle 4t^3(-4sin(t^2)+3cos(t^2)+4sin(t^2)-3cos(t^2))-2t(4cos(t^2)+3sin(t^2))$

$\displaystyle =-2t(4cos(t^2)+3sin(t^2))=-\frac{dy}{dt}$

I did it a different way to how I did before and I now have this answer. I can't see what i'm doing wrong or if it's a typo on the homework sheet.
Yeah, the solution would rather be $\displaystyle t \frac{d^2y}{dt^2}+4t^3 y=0$ but that would be weird because you can simplify by t !

5. Practically every question on the proof by induction homework was a typo. It's very annoying!

I'll have to make a note of this question to quiz my tutor about. There are two questions I have to do in this way, this is the second but the first one is also a typo!!

Someone should proof-read these before they are handed out. It's confusing undergraduate students, namely me!!!