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(b.) Consider the ordinary differential equation in (t > 0) ty" – (t+1)y' + y =0 Note that Y1(t) = et is a solution of this equation. (i). Write y(t) = v(t)e* and derive a differential equation v(t). Solution: • +(1 – ) • = 0 (ii). Solve the equation for v(t) you derive in (a). Solution. • c1( exp( )+ exp(-t)) + c2 (iii). Write out an expression for the general solution of the original ordinary differential equation. Solution: The general solution for y is y(t) = c1( )+c2 exp( ).

(b.) Consider the ordinary differential equation in (t > 0) ty" – (t+1)y' + y =0 Note that Y1(t) = et is a solution of this equation. (i). Write y(t) = v(t)e* and derive a differential equation v(t). Solution: • +(1 – ) • = 0 (ii). Solve the equation for v(t) you derive in (a). Solution. • c1( exp( )+ exp(-t)) + c2 (iii). Write out an expression for the general solution of the original ordinary differential equation. Solution: The general solution for y is y(t) = c1( )+c2 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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(b.) Consider the ordinary differential equation in (t > 0) ty" – (t+1)y' + y = 0 Note that Y1(t) = et is a solution of this equation. (i). Write y(t) = v(t)e* and derive a differential equation v(t). Solution: • +(1 – ) : = 0 (ii). Solve the equation for v(t) you derive in (a). Solution. exp( )+ exp(-t)) + c2 v = (iii). Write out an expression for the general solution of the original ordinary differential equation. Solution: The general solution for y is y(t) = c1 ( )+cz exp( ). (c). Using the method of undetermined coefficients, find the particular solution of the following initial value problem: (a). y" + 4y' + 4y = 2te 24, y(0) = 1, y'(0) = –3 General Solutions: y(t) = c1 exp( +c2 exp(-2t) + 5 exp( Applying initial conditions c1 C2 Particular Solutions: y = exp( )-t exp( )+t3 exp(
Transcribed Image Text:(b.) Consider the ordinary differential equation in (t > 0) ty" – (t+1)y' + y = 0 Note that Y1(t) = et is a solution of this equation. (i). Write y(t) = v(t)e* and derive a differential equation v(t). Solution: • +(1 – ) : = 0 (ii). Solve the equation for v(t) you derive in (a). Solution. exp( )+ exp(-t)) + c2 v = (iii). Write out an expression for the general solution of the original ordinary differential equation. Solution: The general solution for y is y(t) = c1 ( )+cz exp( ). (c). Using the method of undetermined coefficients, find the particular solution of the following initial value problem: (a). y" + 4y' + 4y = 2te 24, y(0) = 1, y'(0) = –3 General Solutions: y(t) = c1 exp( +c2 exp(-2t) + 5 exp( Applying initial conditions c1 C2 Particular Solutions: y = exp( )-t exp( )+t3 exp(
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