Answer to Question #112064 in Differential Equations for Abhijeet Pundir

Question

Solve the ordinary differential equation
Y"+y''+y=0

Answer 1

The question on this form is incorrect if the first term is y''' , the auxiliary equation does not has solution . So i will solve the equation on the form

$y''+y'+y=0$

Set up and solve the auxiliary equation

$\begin{aligned}m^2+m+1&=0\\ \end{aligned}$

Then

$m=\frac{-1\pm \sqrt{1-4}}{2}=\frac{-1}{2}\pm \frac{\sqrt{3}}{2}i$

Here $\alpha =-1/2,\ \ \ \beta= \sqrt{3}/2$ , then the solution given by

$\begin{aligned} y&=e^{\alpha x} [c_1\cos \beta x+c_2\sin \beta x]\\ &= e^{(-1/2)x}[c_1\cos (\sqrt{3}/2) x+c_2\sin(\sqrt{3}/2) x] \end{aligned}$

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If the equation written true as

$y''+y''+y=0\ \ \to 2y''+y=0$

Set up and solve the auxiliary equation

$2m^2+1=0$

Then $m=\pm \frac{1}{\sqrt{2}}$ , here $\alpha =0,\ \ \beta = 1/\sqrt{2},$ then the solution given by

$\begin{aligned} y&=e^{\alpha x} [c_1\cos \beta x+c_2\sin \beta x]\\ &= [c_1\cos (1/\sqrt{2}) x+c_2\sin(1/\sqrt{2}) x] \end{aligned}$