Find the solution of y”+6y’+9=0, y(0)=2, y'(0)=1

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Find the solution of y&#8221;+6y&#8217;+9=0, y(0)=2, y'(0)=1

Suresh Chand · Accepted Answer

Solution Given that y'' + 6y' + 9 = 0                ......(i) with y(0) = 2, y'(0) = 1   ......(ii) Clearly (i) is linear differential equation of second order. The auxiliary equation of (i) is or,   m2 + 6m + 9 = 0 or,   (m + 3)2 = 0 or,   m = -3, -3 Here m has repeated real roots. so the general solution of (i) is y = (c1  + xc2)e-3x                    ..... (iii) y = c1e-3x  + xc2e-3x Differentiating we get (y^{'} = -3c_{1}e^{-3x} + c_{2}e^{-3x} - 3c_{2}xe^{-3x}) y' = c2e-3x - 3(c1+xc2)e-3x             ....(iv) By (ii) we have y(0) = 2 then (iii) gives or,   c1 = 2 Also by (ii) we have y'(0) = 1 then (iv) gives or,   1 = c2e-3.o - 3(c1 + 0.c2)e-3.0 or,   1 = c2 - 3(c1) or,   c2 = 7 Therefore, (iii) becomes u = (2 + 7x)e-3x This is the solution of (i) satisfying (ii)

Find the solution of y”+6y’+9=0, y(0)=2, y'(0)=1

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Find the solution of y”+6y’+9=0, y(0)=2, y'(0)=1

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