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A question regarding initial values and verifying solutions

nathancurtis11

New member
Sep 3, 2013
10
I want to start out with a quick disclaimer, we had a 75 question homework packet assigned a few weeks ago with a few questions from every lecture and this first one is due tomorrow. I missed a lecture, so am completely lost on 3 questions from that lecture. Just dont want it to seem like I'm dumping my whole homework assignment on here so I don't have to do it myself! Just so close to finishing this monstrous packet and need some guidance! Here is the 3rd question:

Question 3:
Let f : (0, +∞) → (all real) be a 2-times differentiable function. Define g: (all real) → (all real) by setting g(t) = f(et).
a) why is g 2-times differentiable
b) Compute the derivatives f′ and f′′ by means of g' and g′′
c). Prove that f is a solution of the differential equation x2y′′+3xy′+y=x if and only if g is a solution of second order differential equation to determine.
 

Chris L T521

Well-known member
Staff member
Jan 26, 2012
995
I want to start out with a quick disclaimer, we had a 75 question homework packet assigned a few weeks ago with a few questions from every lecture and this first one is due tomorrow. I missed a lecture, so am completely lost on 3 questions from that lecture. Just dont want it to seem like I'm dumping my whole homework assignment on here so I don't have to do it myself! Just so close to finishing this monstrous packet and need some guidance! Here is the 3rd question:

Question 3:
Let f : (0, +∞) → (all real) be a 2-times differentiable function. Define g: (all real) → (all real) by setting g(t) = f(et).
a) why is g 2-times differentiable
It should be clear that since $e^t$ is differentiable and $f$ is two times differentiable, then $g^{\prime}(t) = e^tf^{\prime}(e^t)$ and thus $g^{\prime\prime}(t) = e^tf^{\prime}(e^t)+e^{2t}f^{\prime\prime}(e^t)$ by chain rule (it should be clear now as to why we needed $f$ to be twice differentiable). Thus, $g$ is twice differentiable.

b) Compute the derivatives f′ and f′′ by means of g' and g′′
I'll do $f^{\prime}$ and leave $f^{\prime\prime}$ to you. We know that $g^{\prime}(t) = e^t f^{\prime}(e^t)$. Thus, $f^{\prime}(e^t)= e^{-t}g^{\prime}(t)$. Let's make a change of variables, say $s=e^t\implies \ln s= t$. Then it follows that $f^{\prime}(s) = \dfrac{g^{\prime}(\ln s)}{s}$, which is what I think they're looking for (please post back if that's not the case). The calculation for $f^{\prime\prime}$ by means of $g^{\prime\prime}$ should be similar.

c). Prove that f is a solution of the differential equation x2y′′+3xy′+y=x if and only if g is a solution of second order differential equation to determine.
I need some clarification on the part in bold (the "to determine" bit is really throwing me off). Are we showing that $f$ and $g$ are solutions to the same differential equation or are we showing that $f$ is the solution to the differential equation iff $g$ is a solution to some other equation?

Either way, I hope my explanations for (a) and (b) make sense!
 

Opalg

MHB Oldtimer
Staff member
Feb 7, 2012
2,703
I want to start out with a quick disclaimer, we had a 75 question homework packet assigned a few weeks ago with a few questions from every lecture and this first one is due tomorrow. I missed a lecture, so am completely lost on 3 questions from that lecture. Just dont want it to seem like I'm dumping my whole homework assignment on here so I don't have to do it myself! Just so close to finishing this monstrous packet and need some guidance! Here is the 3rd question:

Question 3:
Let f : (0, +∞) → (all real) be a 2-times differentiable function. Define g: (all real) → (all real) by setting g(t) = f(et).
a) why is g 2-times differentiable
b) Compute the derivatives f′ and f′′ by means of g' and g′′
c). Prove that f is a solution of the differential equation x2y′′+3xy′+y=x if and only if g is a solution of second order differential equation to determine.
For c), let $x = e^t$ and let $u(t) = y(x) = y(e^t)$. Using part b) and Chris's comment above, you know that $u'(t) = e^ty'(e^t) = xy'(x)$, and a similar expression for $u''(t)$ in terms of $x^2y''$ and $xy'$. You can then transform the given differential equation (for $y$ as a function of $x$) into a differential equation for $u$ as a function of $t$. Solve that equation, getting $u$ as a function of $t$, and then transform back to get $y$ as a function of $x$.