MHB 2 questions regarding initial values and verifying solutions

[nathancurtis111]

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 don't 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 are the first 2 questions:

Question 1:
The function f: (all real) -> (all real) is defined by the set
{ y''+(pi)²y=0
S: { y(1/3) = (3)^.5
{ y'(1/3) = -pi

Find two numbers (c₁ and c₂) such that
c₁cos((pi)x) + c₂ sin((pi(x))Question 2:
Let f: I -> (all real) be a solution of the diff eq y''-xy'+y=0
a) is f 3 times differentiable?
b) is f smooth?

 

[Chris L T521]

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 don't 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 are the first 2 questions:

Question 1:
The function f: (all real) -> (all real) is defined by the set
{ y''+(pi)²y=0
S: { y(1/3) = (3)^.5
{ y'(1/3) = -pi

Find two numbers (c₁ and c₂) such that
c₁cos((pi)x) + c₂ sin((pi(x))

Since the general solution of $y^{\prime\prime}+\pi^2 y=0$ is $y=c_1\cos(\pi x)+c_2\sin(\pi x)$, it follows that $y^{\prime}=-\pi c_1\sin(\pi x) + \pi c_2\cos(\pi x)$. At this point, you would want to plug in the initial conditions $y(1/3)=\sqrt{3}$ and $y^{\prime}(1/3)=-\pi$ to get the system of equations
\[\left\{\begin{aligned} c_1\cos\left(\frac{\pi}{3}\right) + c_2\sin\left(\frac{\pi}{3}\right) &= \sqrt{3} \\ -\pi c_1\sin\left(\frac{\pi}{3}\right) + \pi c_2\cos\left(\frac{\pi}{3}\right) &= -\pi\end{aligned}\right.\]
I'll leave simplifying the system of equations to you, as well as solving the system. All in all, it shouldn't be too difficult to finish off the problem from here.

Question 2:
Let f: I -> (all real) be a solution of the diff eq y''-xy'+y=0
a) is f 3 times differentiable?
b) is f smooth?

If $f:I\rightarrow \mathbb{R}$ is a solution to $y^{\prime\prime}-xy^{\prime}+y=0$, then we know for sure it's at least two times differentiable; in particular, if $y=f(x)$ is the solution, then we know that $f^{\prime\prime}(x)= xf^{\prime}(x) - f(x)$. Now, $x$, $f(x)$ are at least twice differentiable and $f^{\prime}(x)$ is at least once differentiable; thus, it follows that
\[\frac{d}{dx}\left(xf^{\prime}(x)-f(x)\right)= f^{\prime}(x) + xf^{\prime\prime}(x) - f^{\prime}(x) = xf^{\prime\prime}(x) = x^2f^{\prime}(x)-xf(x).\]
Thus, we've expressed the third derivative of $f$ in terms of functions that are at least once and twice differentiable. To me, this is good enough to show that $f$ is at least three times differentiable. You can extend this argument to showing that $f(x)$ is smooth (i.e. infinitely times differentiable) by showing that the higher order derivatives can be defined in terms of the lower order derivatives.

If you have any follow-up questions, don't hesitate to post them!

I hope this makes sense!

 

[nathancurtis111]

Thank you Chris! Once explained I realized how easy they actually were, just was frustrated never seeing anything of the sort before that wasn't quite sure how to get started.

 

[Mathwebster]

Let me add a bit. As Chris mentioned differentiating both sides gives

$f''' = x f''$

This you can integrate to find $f$ explicitly!

 

[chisigma]

Question 2:
Let f: I -> (all real) be a solution of the diff eq y''-xy'+y=0
a) is f 3 times differentiable?
b) is f smooth?

The solving procedure for a second order incomplete linear ODE ...

$\displaystyle y^{\ ''} - x\ y^{\ '} + y =0\ (1)$

... has been illustrated in...

http://mathhelpboards.com/differential-equations-17/how-solve-differential-equation-second-order-linear-variable-coefficient-2089.html#post9571

If u and v are solution of (1), then is...

$\displaystyle u^{\ ''} - x\ u^{\ '} + u = 0$

$\displaystyle v^{\ ''} - x\ v^{\ '} + v = 0\ (2)$

... and multiplying the first equation by v and the second by u a computing the difference we have...

$\displaystyle v\ u^{\ ''} - u\ v^{\ ''} - x\ (v\ u^{\ '} - u\ v^{\ '}) = 0\ (3)$

... and taking $\displaystyle z= v\ u^{\ '} - u\ v^{\ '}$ we have the first order ODE...

$\displaystyle z^{\ '} = x\ z\ (4)$

... the solution of which is...

$\displaystyle z= c_{2}\ e^{\frac{x^{2}}{2}}\ (5)$

From (5) we3 derive...

$\displaystyle \frac{z}{v^{2}} = \frac{d}{dx} (\frac{u}{v}) = c_{2}\ \frac{e^{\frac{x^{2}}{2}}}{v^{2}} \implies u = c_{1}\ v + c_{2}\ v\ \int \frac{e^{\frac{x^{2}}{2}}}{v^{2}}\ dx\ (6)$

It is easy to verify that $\displaystyle v=x$ is solution of (1) and that means that from (6) we derive that $\displaystyle u = x\ \int \frac{e^{\frac{x^{2}}{2}}}{x^{2}}\ dx$ is also solution and the general solution of (1) is...

$\displaystyle y = c_{1}\ x + c_{2}\ x\ \int \frac{e^{\frac{x^{2}}{2}}}{x^{2}}\ dx\ (7)$

A precise characterization of the u(x) has to be made before to answer the points 1 and 2...

Kind regards

$\chi$ $\sigma$

 

[chisigma]

The solving procedure for a second order incomplete linear ODE ...

$\displaystyle y^{\ ''} - x\ y^{\ '} + y =0\ (1)$

... has been illustrated in...

http://mathhelpboards.com/differential-equations-17/how-solve-differential-equation-second-order-linear-variable-coefficient-2089.html#post9571

If u and v are solution of (1), then is...

$\displaystyle u^{\ ''} - x\ u^{\ '} + u = 0$

$\displaystyle v^{\ ''} - x\ v^{\ '} + v = 0\ (2)$

... and multiplying the first equation by v and the second by u a computing the difference we have...

$\displaystyle v\ u^{\ ''} - u\ v^{\ ''} - x\ (v\ u^{\ '} - u\ v^{\ '}) = 0\ (3)$

... and taking $\displaystyle z= v\ u^{\ '} - u\ v^{\ '}$ we have the first order ODE...

$\displaystyle z^{\ '} = x\ z\ (4)$

... the solution of which is...

$\displaystyle z= c_{2}\ e^{\frac{x^{2}}{2}}\ (5)$

From (5) we3 derive...

$\displaystyle \frac{z}{v^{2}} = \frac{d}{dx} (\frac{u}{v}) = c_{2}\ \frac{e^{\frac{x^{2}}{2}}}{v^{2}} \implies u = c_{1}\ v + c_{2}\ v\ \int \frac{e^{\frac{x^{2}}{2}}}{v^{2}}\ dx\ (6)$

It is easy to verify that $\displaystyle v=x$ is solution of (1) and that means that from (6) we derive that $\displaystyle u = x\ \int \frac{e^{\frac{x^{2}}{2}}}{x^{2}}\ dx$ is also solution and the general solution of (1) is...

$\displaystyle y = c_{1}\ x + c_{2}\ x\ \int \frac{e^{\frac{x^{2}}{2}}}{x^{2}}\ dx\ (7)$

A precise characterization of the u(x) has to be made before to answer the points 1 and 2...

If we analyse the function...

$\displaystyle u(x) = x\ \int \frac{e^{\frac{x^{2}}{2}}}{x^{2}}\ d x\ (1)$

... using the series expansion...

$\displaystyle e^{\frac{x^{2}}{2}} = 1 + \frac{x^{2}}{2} + \frac{x^{4}}{8} + \frac{x^{6}}{48} + ...\ (2)$

... with symple steps we obtain...

$\displaystyle u(x) = -1 + \frac{x^{2}}{2} + \frac{x^{4}}{24} + \frac{x^{6}}{240} + ...\ (3)$

... and the series (3) converges for any real [and complex...] x...

Kind regards

$\chi$ $\sigma$