Second order differential equation in function of a real parameter.

Question

I have this second order ODE: $$X''+2X'=\lambda X$$ With $\lambda \in \mathbb{R}$.

I suppose I have to define three cases:

• $\lambda<0$
• $\lambda=0$
• $\lambda>0$

Could someone give me a hint to start? Thank you :).

Answer 1

As you say, you can distinguish three cases. Start with $\lambda = 0$. You have $x''+2x'=0$. Then you associate to this differential equation the algebraic equation $\mu^2+2\mu = 0$, which gives $\mu = 0$ or $\mu = -2$. Then the solution is $x(t) = c_1+c_2e^{-2t}$. This is a general method for linear homogeneous ODEs and the possible solutions are classified. You can apply this method again for the other cases (see here at the voice Second-order case).

Answer 2

*Just a hint *

$$X''+2X'=\lambda X$$ $$X''+2X'-\lambda X=0 $$

First you need the characteristic polynomial. Here it's

$$R^2+2R-\lambda=0$$ The discriminant is $$\Delta=b^2-4ac=4+4\lambda= 4(1+\lambda)$$ Thats what you have to discuss. Whether the $\Delta$ is 0, >0,<0...

You have three important cases $\lambda=-1 ,\quad \lambda >-1$ and finaly $\lambda<-1$....

Answer 3

The solution rests on the roots of the characteristic equation

$$x^2+2x-\lambda=0$$ which are

$$-1\pm\sqrt{1+\lambda}.$$

For $\lambda>-1$ the two roots are real and for $\lambda>0$, one of them is positive, giving an unstable solution.

For $\lambda<-1$, there are two conjugate roots and a damped oscillatory behavior.