Question

In Problems $\mathbf{37}\mathbf{-}\mathbf{40}$use the concept that $\mathit{y}\mathbf{=}\mathit{c}\mathbf{,}\mathbf{-}\mathbf{\infty }\mathbf{<}\mathit{x}\mathbf{<}\mathbf{\infty }$, is a constant function if and only if $\mathit{y}\mathbf{\text{'}}\mathbf{=}\mathbf{0}$to determine whether the given differential equation possesses constant solutions.

$\mathit{y}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{+}\mathbf{4}\mathit{y}\mathbf{\text{'}}\mathbf{+}\mathbf{6}\mathit{y}\mathbf{=}\mathbf{10}$

Short answer

There exists a constant solution (i.e.) $y=\frac{5}{3}$.

Step-by-step solution

Define constant solution of the function.

When the derivative of a differential equation is zero, the constant solutions appear. If $g\left(a\right)=0$ for some $a$, then $y\left(t\right)=a$is a constant solution of the equation, because $y=f\left(t\right)g\left(a\right)=0$.

Determine whether it has constant solution.

Since$y=c$, implies that $y\text{'}=0$, then the differential equation becomes,

$\begin{array}{rcl}\left(0\right)+4\left(0\right)+6\left(c\right)& =& 10\\ 6c& =& 10\\ c& =& \frac{5}{3}\end{array}$

Hence, the differential equation possesses a constant solution because the value of $c$satisfies the differential equation.

So, the value is $y=\frac{5}{3}$.