Dcs examples for finding the particular solution if it is a root of characteristics equation
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t’s time to start solving constant coefficient, homogeneous, linear, second order differential equations. So, let’s recap how we do this from the last section. We start with the differential equation.
ay′′+by′+cy=0ay″+by′+cy=0
Write down the characteristic equation.
ar2+br+c=0ar2+br+c=0
Solve the characteristic equation for the two roots, r1r1and r2r2. This gives the two solutions
y1(t)=er1tandy2(t)=er2ty1(t)=er1tandy2(t)=er2t
Now, if the two roots are real and distinct (i.e. r1≠r2r1≠r2) it will turn out that these two solutions are “nice enough” to form the general solution
y(t)=c1er1t+c2er2t
ay′′+by′+cy=0ay″+by′+cy=0
Write down the characteristic equation.
ar2+br+c=0ar2+br+c=0
Solve the characteristic equation for the two roots, r1r1and r2r2. This gives the two solutions
y1(t)=er1tandy2(t)=er2ty1(t)=er1tandy2(t)=er2t
Now, if the two roots are real and distinct (i.e. r1≠r2r1≠r2) it will turn out that these two solutions are “nice enough” to form the general solution
y(t)=c1er1t+c2er2t
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