(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( ).

Calculus For The Life Sciences
2nd Edition
ISBN:9780321964038
Author:GREENWELL, Raymond N., RITCHEY, Nathan P., Lial, Margaret L.
Publisher:GREENWELL, Raymond N., RITCHEY, Nathan P., Lial, Margaret L.
Chapter11: Differential Equations
Section11.1: Solutions Of Elementary And Separable Differential Equations
Problem 15E: Find the general solution for each differential equation. Verify that each solution satisfies the...
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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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