# What is the solution of y'' - 4y' + 3y = 0, y(0) = 2, y'(0) = 3?

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Find the solution of the initial value problem: y'' - 4y' + 3y = 0, y(0) = 2, y'(0) = 3

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##### Solution Summary

This shows how to find the general solution for the equation.

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The characteristic equation is given as ar^2 + br + c = 0

or r^2 - 4r + 3 = 0

or (r-1)(r-3) = 0

Therefore, r1 = 1, r2 = ...

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