Find the solution of the following Second Order Differential Equationy″−9y=0,y(0)=2,y′(0)=0

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Mason Hall · Accepted Answer

We have giveny9y=0Auxiliary equation of given differential equationm2−9=0(m−3)(m+3)=0m=3,−3Therefore, general solution of differential equationy=c1e3x+c2e−3xWe have y(0)=2⇒y(0)=c1e3(0)+c2e−3(0)2=c1+c2 (i)y′(x)=3c1e3x−3c2e−3xWe have y′(0)=0⇒y′(x)=3c1e3(0)−3c2e−3(0)0=3c1−3c20=c1−c2 (2)Now, solve equation (1) and (2) we getc1=1,c2=1Therefore, particular solution isy=e3x+e−3x

Linda Birchfield · Accepted Answer

y9y=0,y(0)=2,y′(0)=0The auxiliary equation for this equation is:m2−9=0On solving this equation for &quot;m&quot;.m2−9=0m2=9m=±9m=±3m=−3,3Then, the solution of this given differential equation is:y=c1e−3x+c2e3xGiven that,y(0)=2So, replace x with 0 and y with 2 in it.y=c1e−3x+c2e3x2=c1e−3(0)+c2e3(0)2=c1+c2 (1)Now, differentiate the obtained solution with respect to x.y′=−3c1e−3x+3c2e3xAnd, also given that,y′(0)=0So, replace x with 0 and y with 0 in it.y′=−3c1e−3x+3c2e3x0=−3c1e−3(0)+3c2e3(0)0=−3c1+3c2 (2)Now, solve equations (1) and (2) by the elimination or substitution method.From equation (2), c1=c2, put this value in equation (1), which gives c1=c2=1Then, c1=1,c2=1Thus, the solution of the given differential equation is:y=c1e−3x+c2e3xy=1e−3x+1e3xy=e−3x+e3x

karton · Accepted Answer

y″−9y=0m2−9=0(m−3)(m+3)=0m=3,−3y=c1e3x+c2e−3xy(0)=2⇒y(0)=c1e3(0)+c2e−3(0)2=c1+c2(i)y′(x)=3c1e3x−3c2e−3xy′(0)=0⇒y′(x)=3c1e3(0)−3c2e−3(0)0=3c1−3c20=c1−c2(2)c1=1,c2=1y=e3x+e−3x

Find the solution of the following Second Order Differential Equationy"-9y=0,

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