Find Solutions for Char. Eqtn. y'' + 2y' + 3y = 0 with y(0) = 0, y'(0) = 1

[ognik]

y'' + 2y' + 3y = 0, y(0) = 0, y'(0) = 1

Char. Eqtn. is $p^2 + 2p + 3 = 0, \therefore p = 1 \pm i \sqrt{2}$
Solutions of the form $p=r \pm iq$ are $y = e^{rx} \left( C_1 Cos qx + c_2 Sin qx \right) $

$\therefore y = e^{x} \left( C_1 Cos qx + c_2 Sin qx \right) , q= \sqrt2$

Now $y' = e^{x}\left( - C_1q Sin qx + C_2q Cos qx + C_1 Cos qx + C_2 Sin qx \right)$

$ y(0) = 0 = C_1, y'(0) = 1 = C_2q + C_1, \therefore C_2 = \frac{1}{\sqrt 2}$

Is that right please?

 

[MarkFL]

You cite as the solution to the given second order IVP:

\(\displaystyle y(x)=\frac{1}{\sqrt{2}}e^x\sin\left(\sqrt{2}x\right)\)

I would check first to see if it satisfies the ODE, and so we must compute the first and second derivatives:

\(\displaystyle y'(x)=\frac{1}{\sqrt{2}}\left(\sqrt{2}e^x\cos\left(\sqrt{2}x\right)+e^x\sin\left(\sqrt{2}x\right)\right)=\frac{e^x}{\sqrt{2}}\left(\sqrt{2}\cos\left(\sqrt{2}x\right)+\sin\left(\sqrt{2}x\right)\right)\)

\(\displaystyle y''(x)=\frac{e^x}{\sqrt{2}}\left(\sqrt{2}\cos\left(\sqrt{2}x\right)-2\sin\left(\sqrt{2}x\right)\right)+\frac{e^x}{\sqrt{2}}\left(\sqrt{2}\cos\left(\sqrt{2}x\right)+\sin\left(\sqrt{2}x\right)\right)=\frac{e^x}{\sqrt{2}}\left(2\sqrt{2}\cos\left(\sqrt{2}x\right)-\sin\left(\sqrt{2}x\right)\right)\)

Now, substitution into the ODE gives:

\(\displaystyle \frac{e^x}{\sqrt{2}}\left(2\sqrt{2}\cos\left(\sqrt{2}x\right)-\sin\left(\sqrt{2}x\right)\right)+2\cdot\frac{e^x}{\sqrt{2}}\left(\sqrt{2}\cos\left(\sqrt{2}x\right)+\sin\left(\sqrt{2}x\right)\right)+3\frac{1}{\sqrt{2}}e^x\sin\left(\sqrt{2}x\right)=0\)

Let's multiply through by \(\displaystyle \sqrt{2}e^{-x}\ne0\):

\(\displaystyle \left(2\sqrt{2}\cos\left(\sqrt{2}x\right)-\sin\left(\sqrt{2}x\right)\right)+2\left(\sqrt{2}\cos\left(\sqrt{2}x\right)+\sin\left(\sqrt{2}x\right)\right)+3\sin\left(\sqrt{2}x\right)=0\)

Oops...it appears we have a problem here, because we do not get an identity. Can you find where you made a minor slip?

 

[ognik]

Thanks, my obvious (& silly) 1st mistake was I dropped a - sign, so it should be $ p = -1 \pm i \sqrt 2 $

I followed this through to get $ y = \frac{1}{q}e^{-x} Sin qx $

I then substituted it back into the original eqtn which worked out :-). Thanks for reminding me I could substitute back to check my answer
------------

The next exercise says "Find the general solution of the ODE in the previous question" (i.e. the one we've just looked at). I thought I had found the general solution, what am I missing please?

 

[MarkFL]

Thanks, my obvious (& silly) 1st mistake was I dropped a - sign, so it should be $ p = -1 \pm i \sqrt 2 $

I followed this through to get $ y = \frac{1}{q}e^{-x} Sin qx $

I then substituted it back into the original eqtn which worked out :-). Thanks for reminding me I could substitute back to check my answer
------------

The next exercise says "Find the general solution of the ODE in the previous question" (i.e. the one we've just looked at). I thought I had found the general solution, what am I missing please?

You found the particular solution that satisfies both the given ODE and the initial conditions...you found the solution to the IVP. Now they are asking for the general solution to the ODE, which you already have...the two-parameter family of functions...(containing $c_1$ and $c_2$). :)

 

[ognik]

Thats what I thought, but it seemed a trivial exercise to just write down something I had already found...

Incidentally, I found just 1 sltn for the IVP - I take it there is no 2nd solution to that IVP? For example I remember reading somewhere that if f(x) is a real sltn to an ODE, so is f(-x)? - but I can't remember any qualifiers ...

 

[MarkFL]

A second order linear/homogeneous ODE will have 2 linearly independent solutions. In this case, they are:

\(\displaystyle y_1(x)=c_1e^{-x}\sin\left(\sqrt{2}x\right)\)

\(\displaystyle y_2(x)=c_2e^{-x}\cos\left(\sqrt{2}x\right)\)

And then by the principle of superposition, they are combined to give the general solution. The reasoning is that since both:

\(\displaystyle y_1''+2y_1'+3y_1=0\)

\(\displaystyle y_2''+2y_2'+3y_2=0\)

are true, we can then add and rearrange to obtain:

\(\displaystyle \left(y_1''+y_2''\right)+2\left(y_1'+y_2'\right)+3\left(y_1+y_2\right)=0\)

And given the linearity of differentiation, that is \(\displaystyle \left(f_1+f_2\right)^{(n)}=f_1^{(n)}+f_2^{(n)}\), we may then state that:

\(\displaystyle y(x)=y_1(x)+y_2(x)\)

must also be a solution, and since there are no other linearly independent solutions, so it is the general solution. :)

 

[ognik]

Thanks, so if I find just 1 sltn for the IVP, I leave it at that?

Did you mean $C_2$ for $Y_2$?

 

[MarkFL]

Thanks, so if I find just 1 sltn for the IVP, I leave it at that?

Did you mean $C_2$ for $Y_2$?

Well, you will either be asked to find the general solution, or a particular solution given initial values.

And yes, I had a couple of typos before that I have corrected. :D