# Solve: y''+2y'-3y=0

Solve: $y{}^{″}+2{y}^{\prime }-3y=0$
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sukljama2
${y}^{\prime }=z$
$y{}^{″}=z\frac{dz}{dy}$
$z\frac{dz}{dy}+2z-3y=0$
$zdz+2zdy-3ydy=0$
$zdz=\left(3y-2z\right)dy$
$z=3y-2z$
$z=y$
$y={y}^{\prime }=y{}^{″}$
###### Not exactly what you’re looking for?
puhnut1m
$0=y{}^{″}+2{y}^{\prime }-3y=\left({D}^{2}+2D-3\right)y=\left(D+3\right)\left(D-1\right)y,D=\frac{d}{dx}$
###### Not exactly what you’re looking for?
alenahelenash
Start by finding the characteristic polynomial. In this case it is ${x}^{2}+2x-3$. Find its zeros; they are $x=-3,1$. The general solution to your differential equation is $c{e}^{{x}_{i}t}$, where ${x}_{i}$ is a root to your characteristic polynomial. So we have two linearly independent solutions; $y={e}^{-3t},y={e}^{1t}$. Because this is a linear differential equation, we can take any linear combo of those two solutions, and that will also be a solution. So the general solution for y is $y\left(t\right)={c}_{1}{e}^{-3t}+{c}_{2}{e}^{1t}$where are determined by your initial conditions