Question

In Problemsuse 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.

$\mathbf{3}\mathit{x}\mathit{y}\mathbf{\text{'}}\mathbf{+}\mathbf{5}\mathit{y}\mathbf{=}\mathbf{0}$

Short answer

There exists a constant solution (i.e.) $y=2$.

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,

$\begin{array}{rcl}3x\left(0\right)+5c& =& 10\\ 5c& =& 10\\ c& =& 2\\ & & \end{array}$

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

So, the value is $y=2$.