Question

Find the general solution of the given second-order differential equation $\mathbf{4}\mathit{y}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{+}\mathit{y}\mathbf{\text{'}}\mathbf{=}\mathbf{0}$

Short answer

$y=c_1+c_2{e}^{-\frac{1}{4}x}$

Step-by-step solution

Finding the solution of given second order equation

• A second order differential equation is a type of differential equation that contains a derivative of a function of order 2 and no other higher-order derivatives of the function.
• It contains terms such as y",$\frac{d^{2}y}{d{x}^2}$ , y" (x), and so on that indicate the function's second order derivative.

Solving the second order differential equation

We have the second order differential equation

$4y\text{'}\text{'}+y\text{'}=0······\left(1\right)$

and we have to get its general solution by the following technique:

First, assume that that $y=e^{mx}$is a solution for the homogeneous differential equation Second, differentiate the assumption with respect to, then we have

$y\text{'}=m{e}^{mx}······\left(2\right)$

And

$y^{\text{'}\text{'}}=m^2{e}^{mx}······\left(3\right)$

Substitute with equations (2) and (3) into the differential equation (1), then we obtain

$$\begin{gathered} 4m^2{e}^{mx}+m{e}^{mx}=0 \\ {e}^{mx}\left(4m^2+m\right)=0 \end{gathered}$$

Since $e^{mx}$can not be equal 0 , the we have

$$\begin{gathered} 4m^2+m=0 \\ m(4m+1)=0 \end{gathered}$$

Then the roots are

$m_1=0,and{m}_2=-\frac{1}{4}$

Which are of real and distinct.

Then the general solution of this homogeneous equation$4y\text{'}\text{'}+y\text{'}=0$ can be given as

$$\begin{gathered} y=c_1{e}^{m_{1}x}+c_2{e}^{m_{2}x} \\ =c_1{e}^0+c_2{e}^{-\frac{1}{4}x} \\ =c_1+c_2{e}^{-\frac{1}{4}x} \end{gathered}$$

where $c_1$and $c_2$are an arbitrary constants.

Therefore the general solution is$\mathit{y}\mathbf{=}{\mathit{c}}_{\mathbf{1}}\mathbf{+}{\mathit{c}}_{\mathbf{2}}{\mathit{e}}^{\mathbf{-}\frac{\mathbf{1}}{\mathbf{4}}\mathbf{x}}$