Linda Seales

2021-12-27

Find the solution of the following Differential Equations $y-{y}^{\prime }-6y=0,y\left(0\right)=6,{y}^{\prime }\left(0\right)=13$.

Dabanka4v

Given: $y{y}^{\prime }-6y=0$
with $y\left(0\right)=6,{y}^{\prime }\left(0\right)=13$
we put m for first derivative and ${m}^{2}$ for second derivative then we solve it
$\left({m}^{2}-m-6\right)y=0$
Solving auxilliary equation
${m}^{2}-m-6y=0$
$\left(m+2\right)\left(m-3\right)=0$
Solution is
$y\left(x\right)=A{e}^{-2x}+B{e}^{3x}$
With first initial condition $y\left(0\right)=6$
$6=A+B\dots \left(1\right)$
${y}^{\prime }\left(x\right)=-2A{e}^{-2x}+3B{e}^{3x}$
With second initial condition ${y}^{\prime }\left(0\right)=13$
$-2A+3B=13$ (2)
solving (1) and (2)
$A=1,B=5$
$y={e}^{-2x}+5{e}^{3x}$

Jim Hunt

$y{y}^{\prime }-6y=0$
$y={e}^{kx}$
$\left({e}^{kx}\right){\left({e}^{kx}\right)}^{\prime }-6{e}^{kx}=0$
${k}^{2}{e}^{kx}+k{e}^{kx}-6{e}^{kx}=0$
${e}^{kx}\left({k}^{2}+k-6\right)=0$
${k}^{2}+k-6=0$
$\left(k-2\right)\left(k+3\right)=0$
${k}_{1}=2;{y}_{1}={e}^{2x}$
${k}_{2}=-3;{y}_{2}={e}^{-3x}$
$Y={C}_{1}{e}^{2x}+{C}_{2}{e}^{-3x}$
$\left\{\begin{array}{l}{C}_{1}+{C}_{2}=3\\ 2{C}_{1}-3{C}_{2}=1\end{array}$
$\left\{\begin{array}{l}{C}_{1}=3-{C}_{2}\\ 2{C}_{1}-3{C}_{2}=1\end{array}$
$\left\{\begin{array}{l}{C}_{1}=3-{C}_{2}\\ 2\left(3-{C}_{2}\right)-3{C}_{2}=1\end{array}$
$\left\{\begin{array}{l}{C}_{1}=3-{C}_{2}\\ 6-2{C}_{2}-3{C}_{2}=1\end{array}$
$\left\{\begin{array}{l}{C}_{1}=3-{C}_{2}\\ -5{C}_{2}=-5\end{array}$
$\left\{\begin{array}{l}{C}_{1}=3-{C}_{2}\\ {C}_{2}=1\end{array}$
$\left\{\begin{array}{l}{C}_{1}=2\\ {C}_{2}=1\end{array}$
$Y=2{e}^{2x}+{e}^{-3x}$

karton