find the general solution to the following homogeneous differential equation.
$$100\frac{dy^2}{dx^2} + y = 0$$
Is this worked out by using the auxillary equation such that:
$$100m^2 + 1 = 0$$
so $m = \pm i\sqrt{1/100}$ ?
So the general solution would be $y(x) = A cos (1/10) + B sin(1/10)$?
I am not sure if I've gone about this the right way.
$\endgroup$ 43 Answers
$\begingroup$Note you have
$$\frac{d^2y}{dx^2}=-\frac{1}{100}y$$
Well you only get the derivative proportional to the original function if it is an exponential. Specifically, we have
$$y=e^{\frac{1}{10}ix}$$
Or, equivalently, by Euler's identity,
$$y=A\cos(\frac{1}{10}x)+B\sin(\frac{1}{10}x)$$
so your solution is correct.
$\endgroup$ $\begingroup$As you noted that Auxilliar equation is, $100m^2+1=0$
which gives $$m=\pm\frac{i}{10}$$ i.e. $$m=0\pm\frac{i}{10}$$
Hence the general solution is,
$$y(x)=A\cos(\frac{1}{10})x+B\sin(\frac{1}{10})x$$
$\endgroup$ $\begingroup$Since $$y'' = -\frac{y}{100}$$ Assuming that the solution will be proportional to $e^{\lambda x}$ for some constant $\lambda$. Substitute $y = e^{\lambda x} \implies y'' = \lambda^2 e^{\lambda x}$ into the differential equation to get
$$100\lambda^2 e^{\lambda x} + e^{\lambda x} = 0$$
Dividing through by $e^{\lambda x} \neq 0$ yields $$ \lambda = \frac{i}{10}$$
So our solution is $$y = e^{\frac{i}{10}x} = A \cos \left(\frac{x}{10}\right) + B \sin \left(\frac{x}{10}\right).$$
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