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Linear Algebra With Applications
Found in: Page 441
Linear Algebra With Applications

Linear Algebra With Applications

Book edition 5th
Author(s) Otto Bretscher
Pages 442 pages
ISBN 9780321796974

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Short Answer

Let z1(t)and z2(t) be two complex valued solution of initial valued problem dz(t)dt=λz(t) with z(0)=1 where λ is a complex number. Suppose that z2(t)0for all t.

(a): Using the quotient rule (Exercise 37), show that the derivative oflocalid="1662091749567" style="max-width: none; vertical-align: -20px;" z1(t)z2(t)is zero. Conclude that localid="1662091758671" style="max-width: none; vertical-align: -9px;" z1(t)=z2(t) for all localid="1662091797882" style="max-width: none; vertical-align: -4px;" t .

(b): Show that the initial value problem initial valued problem localid="1662091766773" style="max-width: none; vertical-align: -15px;" dz(t)dt=λz(t) with localid="1662091774693" style="max-width: none; vertical-align: -5px;" z(0)=1 has unique complex-valued solution localid="1662091784628" style="max-width: none; vertical-align: -5px;" z(t).

(a) The derivative of z1(t)z2(t) is 0.

(b) z(t)=eλtiArg(z)

See the step by step solution

Step by Step Solution

(a) Step 1: Determine the derivative of the function z1(t)z2(t).

The derivative of the function g(t)f(t) with respect to t is ddt(g(t)f(t))=f(t)dgdtg(t)dfdt[f(t)]2A.

Consider the complex valued functions z1(t)z2(t).

Using the definition, differentiate z1(t)z2(t) with respect tot as follows.


Step 2: Simplify the equation ddt{z1(t)z2(t)}=z2(t)dz1(t)dt-z1(t)dz2(t)dt[z2(t)]2.

As ddt{z(t)}=λz(t), substitute the values λz1(t) for ddt{z1(t)} and λz2(t) for ddt{z2(t)} in the equation ddt{z1(t)z2(t)}=z2(t)dz1(t)dtz1(t)dz2(t)dt[z2(t)]2 as follows.


Hence, the derivative of z1(t)z2(t) is 0 .

(b)Step 3: Simplify the equation dz(t)dt=λz(t).

Consider the initial value problem dz(t)d(t)=λz(t) with z(0)=1 .

Simplify the equation dz(t)d(t)=λz(t) as follows.


Step 4: Determine the solution.

Take integration both side in the equation dz(t)z(t)=λd(t) as follows.


Substitute the value 0 for t in the equation lnz(t)=λt+c as follows.


Substitute the value 0 for c in the equation lnz(t)=λt+c as follows.


As lnz(t)=ln|z(t)|+iArg(z), Substitute the value ln|z(t)|+iArg(z) for lnz(t)in the equation lnz(t)=λt as follows.


Taking anti-log both side in the equation ln|z(t)|=λtiArg(z) as follows.


Hence, the solution of the initial value problem dz(t)d(t)=λz(t) with z(0)=1 is z(t)=eλtiArg(z).

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