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Fundamentals Of Differential Equations And Boundary Value Problems
Found in: Page 341
Fundamentals Of Differential Equations And Boundary Value Problems

Fundamentals Of Differential Equations And Boundary Value Problems

Book edition 9th
Author(s) R. Kent Nagle, Edward B. Saff, Arthur David Snider
Pages 616 pages
ISBN 9780321977069

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Short Answer

Derive the system (7) in the special case when n = 3. [Hint: To determine the last equation, require that I.[yp]=g and use the fact that y1,y2, and satisfy the corresponding homogeneous equation.]

The three functions V1,V2 and V3 that satisfy the systems are given as;

V1'y1+V2'y2+V3'y3=0V1'y1'+V2'y2'+V3'y3'=0V1'y1''+V2'y2''+V3'y3''=g.

See the step by step solution

Step by Step Solution

Step 1: Determine the three unknown function

Consider the differential equation

y'''(x)+p1y''(x)+p2y'(x)+p3y(x)=g(x)

where the coefficient functions p1,p2 and p3 as well as g are continuous on(a,b) . To find a particular solution to the given equation we need to know a fundamental solution set y1,y2,y3 for the corresponding homogeneous equation

y'''(x)+p1y''(x)+p2y'(x)+p3y(x)=0.

Therefore, a general solution to this homogeneous equation is

yh(x)=c1y1(x)+c2y2(x)+c3y3(x)

where c1,c2and c3 are arbitrary constants. In the method of variation of parameters, we assume there exists a particular solution to the given equation of the form

yp(x)=V1(x)y1(x)+V2(x)y2(x)+V3(x)y3(x)

and we try to determine the functions V1(x),V2(x) and V3(x). There are three unknown functions so we will need three equations to determine them. Differentiating yp(x) gives us

yp'=V1y1'+V2y2'+V3y3'+V1'y1+V2'y2+V3'y3.

To prevent second derivatives of the unknowns V1,V2 and V3 from entering the formula yp'' we impose the condition

V1'y1+V2'y2+V3'y3=0

In the same manner, we impose the next condition

V1'y1+V2'y2+V3'y3=0

Step 2: Determine the solution by using three equations.

Finally, the third condition that we impose is that ypsatisfies the given equation.

yp'''(x)+p1yp''(x)+p2yp'(x)+p3yp(x)=g(x)V1y1'''+V2y2'''+V3y3'''+V1'y1''+V2'y2''+V3'y3''+p1V1y1''+V2y2''+V3y3''+p2V1y1'+V2y2'+V3y3'+p3V1y1+V2y2+V3y3=g(x)V1y1'''+p1y1''+p2y1'+p3y1+V2y2'''+p1y2''+p2y2'+p3y2+V3y3'''+p1y3''+p2y3'+p3y3+V1'y1''+V2'y2''+V3'y3''=g(x)V1'y1''+V2'y2''+V3'y3''=g(x)

So, using the previous conditions and the fact that y1,y2and y3are solutions to the homogenous equation we get;

V1'y1''+V2'y2''+V3'y3''=g.

Therefore, we seek three functions V1,V2 and V3 that satisfy the system;

V1'y1+V2'y2+V3'y3=0V1'y1'+V2'y2'+V3'y3'=0V1'y1''+V2'y2''+V3'y3''=g

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