MHB Limit of the solution - Which is the conclusion?

[mathmari]

Hey!

I am looking at the following exercise:

Let $m_1\neq m_2$ be constants and $y$ the solution of the initial value problem $$y''-(m_1+m_2)y'+m_1m_2y=0\ \ \ \ y(0)=0 \ \ y'(0)=1$$
We consider $y$ as a function not only of $x$ but also of $m_1$ and $m_2$.
With constant $m_2$ find (if exists) the limit of $y$ when $m_1\rightarrow m_2$.
What do we conclude? I have done the following:

$$y''-(m_1+m_2)y'+m_1m_2y=0$$
The characteristic polynomial is $$k^2-(m_1+m_2)k+m_1m_2=0$$
We get the following: $$\Delta=m_1^2+2m_1m_2+m_2^2-4m_1m_2=(m_1-m_2)^2$$
Therefore $$k_{1,2}=\frac{m_1+m_2\pm (m_1-m_2)}{2}$$
So, we get $$k_1=\frac{m_1+m_2+m_1-m_2}{2}=m_1, \ \ k_2=\frac{m_1+m_2-m_1+m_2}{2}=m_2$$

The solution of the ivp is $$y(x)=c_1e^{m_1x}+c_2e^{m_2x}$$

$$y(0)=0 \Rightarrow c_1+c_2=0 \Rightarrow c_1=-c_2$$
$$y'(x)=c_1m_1e^{m_1x}+c_2m_2e^{m_2x} \ : \ y'(0)=1 \Rightarrow c_1m_1+m_2c_2=1 \Rightarrow c_1=\frac{1-c_2m_2}{m_1}$$

The solution of the ivp is therefore, $$y(x)=\frac{1-c_2m_2}{m_1}e^{m_1x}+c_2e^{m_2x}$$

We have that \begin{align*}\lim_{m_1\rightarrow m_2}y(x, m_1, m_2)&=\frac{1-c_2m_2}{m_2}e^{m_2x}+\frac{c_2m_2}{m_2}e^{m_2x} \\ & =\frac{1}{m_2}e^{m_2x}-\frac{c_2m_2}{m_2}e^{m_2x}+\frac{c_2m_2}{m_2}e^{m_2x} \\ & = \frac{e^{m_2x}}{m_2}\end{align*}

Is everything correct? (Wondering)

What could we conclude? (Wondering)

 

[I like Serena]

Hey mathmari! (Smile)

We have that \begin{align*}\lim_{m_1\rightarrow m_2}y(x, m_1, m_2) &= \frac{e^{m_2x}}{m_2}\end{align*}

Is everything correct? (Wondering)

Isn't $y(0) = \frac{e^{m_2x}}{m_2}\Big|_{x=0} = \frac 1{m_2} \ne 0$? (Wondering)

Let's back up a bit:

The solution of the ivp is $$y(x)=c_1e^{m_1x}+c_2e^{m_2x}$$

$$y(0)=0 \Rightarrow c_1+c_2=0 \Rightarrow c_1=-c_2$$
$$y'(x)=c_1m_1e^{m_1x}+c_2m_2e^{m_2x} \ : \ y'(0)=1 \Rightarrow c_1m_1+m_2c_2=1 \Rightarrow c_1=\frac{1-c_2m_2}{m_1}$$
The solution of the ivp is therefore, $$y(x)=\frac{1-c_2m_2}{m_1}e^{m_1x}+c_2e^{m_2x}$$

We have a system of 2 equations in $c_1$ and $c_2$.
Shouldn't we be able to deduce both of them from that? (Wondering)

 

[mathmari]

I tried it again... (Thinking)

The solution of the ivp is $$y(x)=c_1e^{m_1x}+c_2e^{m_2x}$$

$$y(0)=0 \Rightarrow c_1+c_2=0 \Rightarrow c_1=-c_2$$

So, we get $y(x)=-c_2e^{m_1x}+c_2e^{m_2x}$.

The first derivative of $y(x)$ is $y'(x)=-c_2m_1e^{m_1x}+c_2m_2e^{m_2x}$.

From the condition $y'(0)=1$ we get $-c_2m_1+c_2m_2=1 \Rightarrow c_2(m_2-m_1)=1 \Rightarrow c_2=\frac{1}{m_2-m_1}$.

The solution of the ivp is therefore, $$y(x)=-\frac{1}{m_2-m_1}e^{m_1x}+\frac{1}{m_2-m_1}e^{m_2x}=\frac{e^{m_2x}-e^{m_1x}}{m_2-m_1}$$

We have that $$\lim_{m_1\rightarrow m_2}y(x, m_1, m_2)=\lim_{m_1\rightarrow m_2}\frac{e^{m_2x}-e^{m_1x}}{m_2-m_1}=\frac{0}{0}$$

Applying the De L'Hospital Rule we get $$\lim_{m_1\rightarrow m_2}y(x, m_1, m_2)=\lim_{m_1\rightarrow m_2}\frac{-xe^{m_1x}}{-1}=xe^{m_2x}$$ Is everything correct now? (Wondering)

 

[I like Serena]

Applying the De L'Hospital Rule we get $$\lim_{m_1\rightarrow m_2}y(x, m_1, m_2)=\lim_{m_1\rightarrow m_2}\frac{-xe^{m_1x}}{-1}=xe^{m_2x}$$

Is everything correct now? (Wondering)

Let's see... (Thinking)

$y(0)=0$ check.

$y'(0)=e^0 + 0 \cdot m_2 \cdot e^0 = 1$ check.

\begin{aligned}y'' - 2m_2 y' + m_2^2 y
&= \d{}x(e^{m_2x} + m_2 x e^{m_2x}) - 2m_2(e^{m_2x} + m_2 x e^{m_2x}) + m_2^2xe^{m_2x} \\
&= m_2e^{m_2x} + m_2 e^{m_2x} + m_2^2xe^{m_2x} - 2m_2(e^{m_2x} + m_2 x e^{m_2x}) + m_2^2xe^{m_2x} \\
&= 0
\end{aligned}
check.

Yep. I believe it is correct. (Nod)

So what is the conclusion?

 

[mathmari]

Let's see... (Thinking)

$y(0)=0$ check.

$y'(0)=e^0 + 0 \cdot m_2 \cdot e^0 = 1$ check.

\begin{aligned}y'' - 2m_2 y' + m_2^2 y
&= \d{}x(e^{m_2x} + m_2 x e^{m_2x}) - 2m_2(e^{m_2x} + m_2 x e^{m_2x}) + m_2^2xe^{m_2x} \\
&= m_2e^{m_2x} + m_2 e^{m_2x} + m_2^2xe^{m_2x} - 2m_2(e^{m_2x} + m_2 x e^{m_2x}) + m_2^2xe^{m_2x} \\
&= 0
\end{aligned}
check.

Yep. I believe it is correct. (Nod)

(Happy)

So what is the conclusion?

When we have the initial value problem $$y''-2my'+m^2y=0\ \ \ \ y(0)=0 \ \ y'(0)=1$$ the solution will be $xe^{mx}$, right? (Wondering) What else could we conclude? (Wondering)

 

[I like Serena]

When we have the initial value problem $$y''-2my'+m^2y=0\ \ \ \ y(0)=0 \ \ y'(0)=1$$ the solution will be $xe^{mx}$, right? (Wondering)

What else could we conclude? (Wondering)

When we have the ivp $y''-(m_1+m_2)y'+m_1m_2y=0,\ \ \ \ y(0)=0, \ \ y'(0)=1$, the solution for $m_1\ne m_2$ is:
$$y(x) = \frac{e^{m_2x} - e^{m_1x}}{m_2 - m_1}$$
This solution does not exist if $m_1=m_2$.
In that case the solution is:
$$y(x) = xe^{m_2x}$$
which we could find by taking the limit. (Nerd)