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Using the method of undetermined coefficients, find the particular solution of the following initial value problem: (а). у" + 4у' + 4y %3D 2ie , у0) %3D 1, У (0) — -3 = General Solutions: y(t) = c1 exp(| +c2 exp(-21) + еxp( Applying initial conditions c¡ = , C2 = )-t exp( + exp( Particular Solutions: y= exp(

Using the method of undetermined coefficients, find the particular solution of the following initial value problem: (а). у" + 4у' + 4y %3D 2ie , у0) %3D 1, У (0) — -3 = General Solutions: y(t) = c1 exp(| +c2 exp(-21) + еxp( Applying initial conditions c¡ = , C2 = )-t exp( + exp( Particular Solutions: y= exp(

Advanced Engineering Mathematics
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
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Using the method of undetermined coefficients, find the particular solution of the following initial value problem: (a). y" + 4y' + 4y = 2te-2", y(0) = 1, y'(0) = -3 General Solutions: y(t) = c1 exp( exp(-21) + eхp( +c2 Applying initial conditions ci = C2 = exp( )-t exp( )+ exp( Particular Solutions: y = (b). у" — у — у %3D 0, у(0) %3D 0, у (0) %3 3, у"(0) %3 2 General Solutions: y(t) : c1 cos( C2 sin( )+c3 exp Applying initial conditions c1 C2 = C3 = Particular Solutions: y(t) = + cos( sin( еxp
Transcribed Image Text:Using the method of undetermined coefficients, find the particular solution of the following initial value problem: (a). y" + 4y' + 4y = 2te-2", y(0) = 1, y'(0) = -3 General Solutions: y(t) = c1 exp( exp(-21) + eхp( +c2 Applying initial conditions ci = C2 = exp( )-t exp( )+ exp( Particular Solutions: y = (b). у" — у — у %3D 0, у(0) %3D 0, у (0) %3 3, у"(0) %3 2 General Solutions: y(t) : c1 cos( C2 sin( )+c3 exp Applying initial conditions c1 C2 = C3 = Particular Solutions: y(t) = + cos( sin( еxp
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