Question #289483

Show that 𝑦=𝑐1𝑒𝑥+𝑐2𝑒2𝑥 is the general solution of 𝑦′′−3𝑦′+2𝑦=0 on any interval, and find the particular solution for which 𝑦 0 =−1 and 𝑦′(0)=1. 


1
Expert's answer
2022-01-24T18:09:53-0500

Given:

𝑦=𝑐1𝑒𝑥+𝑐2𝑒2𝑥𝑦=𝑐_1𝑒^𝑥+𝑐_2𝑒^{2𝑥}


We have:

y=𝑐1𝑒𝑥+2𝑐2𝑒2𝑥y'=𝑐_1𝑒^𝑥+2𝑐_2𝑒^{2𝑥}

y=𝑐1𝑒𝑥+4𝑐2𝑒2𝑥y''=𝑐_1𝑒^𝑥+4𝑐_2𝑒^{2𝑥}

Hence


𝑦3𝑦+2𝑦=(𝑐1𝑒𝑥+4𝑐2𝑒2𝑥)3(𝑐1𝑒𝑥+2𝑐2𝑒2𝑥)+2(𝑐1𝑒𝑥+𝑐2𝑒2𝑥)=0𝑦''−3𝑦'+2𝑦=(𝑐_1𝑒^𝑥+4𝑐_2𝑒^{2𝑥})\\ -3(𝑐_1𝑒^𝑥+2𝑐_2𝑒^{2𝑥})+2(𝑐_1𝑒^𝑥+𝑐_2𝑒^{2𝑥})=0

The initial conditions give

y(0)=c1+c2=1y(0)=c_1+c_2=-1

y(0)=c1+2c2=1y'(0)=c_1+2c_2=1

Roots:

c1=3,c2=2c_1=-3,\quad c_2=2

Finally

𝑦=3𝑒𝑥+2𝑒2𝑥𝑦=-3𝑒^𝑥+2𝑒^{2𝑥}


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