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Can someone help me with this.. 2y''+7y'+3y+0 .: y'(0)=0, y(0)=3
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Have you considered the Characteristic Equation?
2y''+7y'+3y=0 .: y'(0)=0, y(0)=3...
@ apad12 is ur question 2y''+7y'+3y=0
So, then that's a "no". Given such a problem statement, you should have seen this associated equation, the Characteristic Equation: \(2r^{2} + 7r + 4 = 0\) Your first task is to solve this equation. It is easier if you can factor the LHS.
then the characteristic equation is 2m^2 +7m +3 =0 where y=e^(mx) is a trial solution 2m^2 +6m+1m +3=0 2m(m+3)+1(m+3)=0 (m+3)(2m+1)=0 hence m=-3,-0.5 hence y=Ae^(-3x) + Be^(-0.5x)
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