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.
Given:
𝑦=𝑐1𝑒𝑥+𝑐2𝑒2𝑥𝑦=𝑐_1𝑒^𝑥+𝑐_2𝑒^{2𝑥}y=c1ex+c2e2x
We have:
y′=𝑐1𝑒𝑥+2𝑐2𝑒2𝑥y'=𝑐_1𝑒^𝑥+2𝑐_2𝑒^{2𝑥}y′=c1ex+2c2e2x
y′′=𝑐1𝑒𝑥+4𝑐2𝑒2𝑥y''=𝑐_1𝑒^𝑥+4𝑐_2𝑒^{2𝑥}y′′=c1ex+4c2e2x
Hence
The initial conditions give
Roots:
Finally
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