If roots of the auxiliary equation of $\frac{d^{2}y}{d{x}^2}+a\frac{dy}{dx}+by=0$ are real and equal, the general solution of the differential equation is

Differential Equation General Solution

This question asks us to find the general solution for a given second-order linear homogeneous differential equation with constant coefficients. A crucial piece of information provided is that the roots of its auxiliary equation are real and equal.

Auxiliary Equation Derivation

The given differential equation is:

$\frac{d^{2}y}{d{x}^2}+a\frac{dy}{dx}+by=0$

To solve a homogeneous linear differential equation with constant coefficients, we first form its auxiliary equation (also known as the characteristic equation). This is done by replacing $\frac{d^{2}y}{d{x}^2}$ with $m^2$, $\frac{dy}{dx}$ with $m$, and $y$ with $1$. For our equation, the coefficients are $1$ for $\frac{d^{2}y}{d{x}^2}$, $a$ for $\frac{dy}{dx}$, and $b$ for $y$.

So, the auxiliary equation is:

$m^2+am+b=0$

Roots Condition Analysis

We are told that the roots of this auxiliary equation are real and equal. For a quadratic equation in the form $A{m}^2+Bm+C=0$, the roots are real and equal if and only if its discriminant ($\Delta =B^2-4AC$) is zero.

In our auxiliary equation $m^2+am+b=0$:

- The coefficient $A$ (of $m^2$) is $1$.
- The coefficient $B$ (of $m$) is $a$.
- The constant term $C$ is $b$.

Setting the discriminant to zero:

$a^2-4(1)(b)=0$

$a^2-4b=0$

This implies that $b=\frac{a^2}{4}$. This relationship between $a$ and $b$ ensures the roots are real and equal.

Determining the Repeated Root Value

When a quadratic equation $A{m}^2+Bm+C=0$ has real and equal roots, the single repeated root $m$ can be found using the formula $m=\frac{-B}{2A}$.

Applying this formula to our auxiliary equation $m^2+am+b=0$:

$m=\frac{-a}{2(1)}$

$m=-\frac{a}{2}$

Therefore, the repeated root of the auxiliary equation is $m=-\frac{a}{2}$.

General Solution Formula

For a second-order linear homogeneous differential equation with constant coefficients, if the auxiliary equation yields real and equal roots (let's say $m_1=m_2=m$), the general solution is given by the specific form:

$y=(c_1+c_{2}x)e^{mx}$

Here, $c_1$ and $c_2$ are arbitrary constants that would typically be determined by initial or boundary conditions if provided.

Solution Application and Result

Now, we substitute the calculated repeated root $m=-\frac{a}{2}$ into the general solution formula:

$y=(c_1+c_{2}x)e^{\left(-\frac{a}{2}\right)x}$

$y=(c_1+c_{2}x)e^{-\frac{ax}{2}}$

Comparing Solution with Options

Let's compare our derived general solution with the given options to find the correct match:Based on the analysis, the general solution for the given differential equation when its auxiliary roots are real and equal is $y=(c_1+c_{2}x)e^{-\frac{ax}{2}}$.