MHB Differential Equation challenge

[Petrus]

Hello MHB,
I wanted to post a challange question that is hopefully not really difficult, if the question is not understandable make sure to write it so I can try explain!:)Calculate the Differential equation for
$$y''+2y'=0$$
that satisfy
$$\lim_{x->\infty}y(x)=1$$ and $$y(0)=0$$

Regards,
$$|\pi\rangle$$

 

[MarkFL]

Re: Differential Equation challange

Hello Petrus,

One way to proceed is to recognize we are given a linear second order homogeneous ODE whose characteristic roots are:

$$r=0,-2$$

Hence, the general solution is:

$$y(x)=c_1+c_2e^{-2x}$$

We may now state:

$$\lim_{x\to\infty}y(x)=c_1=1$$

$$y(0)=1+c_2=0\,\therefore\,c_2=-1$$

Thus the solution satisfying the given conditions is:

$$y(x)=1-e^{-2x}$$

 

[Petrus]

Re: Differential Equation challange

Hello Petrus,

One way to proceed is to recognize we are given a linear second order homogeneous ODE whose characteristic roots are:

$$r=0,-2$$

Hence, the general solution is:

$$y(x)=c_1+c_2e^{-2x}$$

We may now state:

$$\lim_{x\to\infty}y(x)=c_1=1$$

$$y(0)=1+c_2=0\,\therefore\,c_2=-1$$

Thus the solution satisfying the given conditions is:

$$y(x)=1-e^{-2x}$$

Hello MarkFL,
Congrats that is the correct answer(Clapping)
Well you said "one way to proceed.." is there more way to solve this differential equation? Well this is the way I have learned yet:) The smart thing with start with the $$\lim_{x->\infty}y(x)=1$$ is that $$\lim_{x->\infty}e^{-2x}=0$$ so that $$c_2$$ can be any real value and hence we are only left with $$c_1$$ and thanks to that we can solve it with $$y(0)=0$$ Thanks for taking your time and enter my challange!:)

Regards,
$$|\pi\rangle$$

 

[MarkFL]

Another way is to write the ODE as:

$$\frac{d}{dx}\left(\frac{dy}{dx} \right)=-2\frac{dy}{dx}$$

Integrate with respect to $x$ to obtain:

$$\frac{dy}{dx}=-2y+c_1$$

Separate variables:

$$\frac{1}{-2y+c_1}\,dy=dx$$

Integrate:

$$-\frac{1}{2}\ln|c_1-2y|=x+c_2$$

$$\ln|c_1-2y|=-2x+c_2$$

$$c_1-2y=c_2e^{-2x}$$

$$y(x)=c_1+c_2e^{-2x}$$

And now proceed as before...

 

[alyafey22]

Another way is to use the substitution $$t = y'$$

$$t'+2t=0$$

This differential equation is separable we can also multiply by the integrating factor .

 

[Chris L T521]

Another method (which I think is the longest way; EDIT: Solving by Power Series about $x=0$ would be way more complicated and longer) would be to take Laplace Transforms of both sides to obtain:
\[\mathcal{L}\{y^{\prime\prime}\} + \mathcal{L}\{2y\} = \mathcal{L}\{0\} \implies s^2Y(s)-sy(0) - y^{\prime}(0) + 2sY(s)-2y(0)=0.\]
Note here we know nothing about $y^{\prime}(0)$, but let's not worry about that for the time being. Letting $y(0)=0$ and solving for $Y(s)$, we get
\[Y(s)=\frac{y^{\prime}(0)}{s^2+2s}=\frac{y^{\prime}(0)(s+2-s)}{2s(s+2)}=\frac{y^{\prime}(0)}{2}\left(\frac{1}{s}-\frac{1}{s+2}\right)\]
Taking the Inverse Laplace Transform of both sides gives us
\[y(x)=\mathcal{L}^{-1}\{Y(s)\} = \frac{y^{\prime}(0)}{2}\left(\mathcal{L}^{-1}\left\{\frac{1}{s}\right\} - \mathcal{L}^{-1}\left\{\frac{1}{s+2}\right\}\right) = \frac{y^{\prime}(0)}{2}(1-e^{-2x})\]
But we're told that $\displaystyle\lim_{x\to\infty}y(x)=1$, which implies that
\[\lim_{x\to\infty}y(x)= \lim_{x\to\infty}\frac{y^{\prime}(0)}{2} (1-e^{-2x})=\frac{y^{\prime}(0)}{2}=1\implies y^{\prime}(0)=2.\]
Therefore, the solution to our equation is $y(x)=1-e^{-2x}$.

(That was fun! XD)

 

[alyafey22]

Since $$y(x) $$ is analytic around $$0$$ we can find the power series

Let $$y(x) = \sum_{n\geq 0} a_n x^n$$

$$y'(x) =\sum_{n\geq 1} n a_n x^{n-1} $$

$$y''(x) =\sum_{n\geq 2} n(n-1) a_n x^{n-2} $$

Hence we have the following

$$\sum_{n\geq 2} n(n-1) a_n x^{n-2} +2 \, \sum_{n\geq 1} n a_n x^{n-1}=0$$

$$\sum_{n\geq 2} n(n-1) a_n x^{n-2} +2 \, \sum_{n\geq 2} (n-1) a_{n-1} x^{n-2}=0$$

Hence we have the following

$$n(n-1)a_n+2(n-1) a_{n-1} =0$$$$na_n+2a_{n-1}=0 \,\,\, \Rightarrow \,\,\, a_n = -\frac{2}{n} a_{n-1} \,\,\,\, , n\geq 2 $$

$$a_2 = -a_1$$

$$a_3 = \frac{2}{3}a_1$$

$$a_4 = -\frac{2^2}{3 \cdot 4} a_1$$

which suggests that

$$a_n = \frac{(-2)^{n-1}}{n!} a_1$$

we can easily deduce that $$a_0 = 0 $$ using the initial condition .

$$y(x) = a_1\sum_{n\geq 1}\frac{(-2)^{n-1}}{(n)!} \, x^n = \frac{-a_1}{2}\, \sum_{n\geq 1}\frac{(-2)^{n}}{(n)!} \, x^n$$

$$y(x) = \frac{a_1}{-2} (e^{-2x}-1) $$

using $$ y(\infty) = 1$$ implies that $$a_1 = 2$$

$$y(x) = -(e^{-2x}-1) = 1-e^{-2x} \,\, \square $$

By the way I didn't see Chris's comment (Rofl)

 

[Petrus]

Hello,
I want to first thank you all for taking your time and enter the challange!:) When I did post this problem I only knew one method (which is the MarkFL post #2) and now you all show me more method! Remainds me one of the reason why I find math is such an intressting subject! I am Really gratefull!

Regards,
$$|\pi\rangle$$