Question

For the second order linear ordinary differential equation.
$\frac{d^{2}y}{d{x}^2}+p\frac{dy}{dx}+qy=0$, where p and q real numbers.
The following function is a solution $y=e^{\lambda x}$
Which one of the following statements is Not TRUE?

• A. $\lambda$ has two values : one complex and one real
• B. ${\lambda }^2+p\lambda +q=0$
• C. $\lambda$ has two real values
• D. $\lambda$ has two complex values

Answer 1

The correct option is A $\lambda$ has two values : one complex and one real

Given

$y=e^{\lambda x}$

$\frac{dy}{dx}=\lambda e^{\lambda x}$

$and\frac{d^{2}y}{d{x}^2}={\lambda }^2.e^{\lambda x}$

Put the values in the given differential equation,

we get ${\lambda }^2.e^{\lambda x}+p\lambda e^{\lambda x}+q{e}^{\lambda x}=0$

$\Rightarrow e^{\lambda x}({\lambda }^2+p\lambda +q)=0$

$\Rightarrow e^{\lambda x}=0and{\lambda }^2+p\lambda +q=0)$ (statement b is correct)

For finding values of $\lambda$

${\lambda }^2+p\lambda +q=0$

$\lambda =\frac{-p\pm \sqrt{p^2-4pq}}{2}$

Here, there are three possibilities

(i) if $p^2-4pq>0$ in this case,$\lambda$ have two values and both are real

(ii) if $p^2-pq=0$ in this case ,$\lambda$ have two real but equal values.

(iii) if $p^2-4pq<0.$ in this case , $\lambda$ have two complex values.

Here, statement b, c and d are correct. But statement a is not true.