# Find the general soluton to the differential equation by first finding the roots of the auxiliary...

## Question:

Find the general soluton to the differential equation by first finding the roots of the auxiliary equation. Pay close attention to the order of each term.

{eq}(a) y'''+4y"=0 \\(b) y'''-4y'=0 \\(c) y"+4y'+5y=0 {/eq}

## Differential Equation:

A differential equation is an equation that contains some function and its one or more derivatives.

A first-order linear, homogenous differential equation is of the form of {eq}ay'+by=0 {/eq}

A second-order linear, homogenous differential equation is of the form of {eq}ay''+by'+cy=0 {/eq}.

A third-order linear, homogenous differential equation is of the form of {eq}ay'''+by''+cy'+d=0 {/eq}.

(a) Given

{eq}y'''+4y"=0 {/eq}

We have to find the general solution of the differential equation.

Put {eq}y=e^{kx}\\ {/eq}

The derivatives are

{eq}y'=ke^{kx}\\ y''=k^{2}e^{kx}\\ y'''=k^{3}e^{kx} {/eq}

Substituting the values in a given differential equation, we have

{eq}y'''+4y''=0\\ k^{3}e^{kx}+4k^{2}e^{kx}=0\\ e^{kx}\left ( k^{3}+4k^{2} \right )=0 {/eq}

Now, the auxiliary equation is

{eq}k^{3}+4k^{2}=0\\ k^{2}(k+4)=0\\ k=0,0.-4 {/eq}

{eq}\therefore {/eq} The general solution of the differential equation is

{eq}\color{blue}{y}=\color{blue}{c_1+c_2x+c_3e^{-4x}} {/eq}

(b) Given

{eq}y'''-4y'=0 {/eq}

We have to find the general solution of the differential equation.

Put {eq}y=e^{kx}\\ {/eq}

The derivatives are

{eq}y'=ke^{kx}\\ y''=k^{2}e^{kx}\\ y'''=k^{3}e^{kx} {/eq}

Substituting the values in a given differential equation, we have

{eq}y'''-4y'=0\\ k^{3}e^{kx}-4ke^{kx}=0\\ e^{kx}\left ( k^{3}-4k \right )=0 {/eq}

Now, the auxiliary equation is

{eq}k^{3}-4k=0\\ k(k^{2}-4)=0\\ k(k^{2}-2^{2})=0\\ k(k+2)(k-2)=0\\ k=0,-2,2 {/eq}

{eq}\therefore {/eq} The general solution of the differential equation is

{eq}\color{blue}{y}=\color{blue}{c_1+c_2e^{-2x}+c_3e^{2x}} {/eq}

(C) Given

{eq}y''+4y'+5y=0 {/eq}

We have to find the general solution of the differential equation.

Put {eq}y=e^{kx}\\ {/eq}

The derivatives are

{eq}y'=ke^{kx}\\ y''=k^{2}e^{kx}\\ {/eq}

Substituting the values in a given differential equation, we have

{eq}y''+4y'+5y=0\\ k^{2}e^{kx}+4ke^{kx}+5e^{kx}=0\\ e^{kx}\left ( k^{2}+4k+5 \right )=0 {/eq}

Now, the auxiliary equation is

{eq}k^{2}+4k+5=0\\ \begin{align} k &=\dfrac{-4\pm \sqrt{(4)^{2}-4(1)(5)}}{2}\\ &=\dfrac{-4\pm \sqrt{16-20}}{2}\\ &=\dfrac{-4\pm \sqrt{-4}}{2}\\ &=\dfrac{-4\pm 2\imath}{2}\\ &=-2\pm \imath \end{align} {/eq}

{eq}\therefore {/eq} The general solution of the differential equation is

{eq}\color{blue}{y}=\color{blue}{e^{-2x}(A\cos x +B \sin x)} {/eq}