• :00Days
  • :00Hours
  • :00Mins
  • 00Seconds
A new era for learning is coming soonSign up for free
Log In Start studying!

Select your language

Suggested languages for you:
Answers without the blur. Sign up and see all textbooks for free! Illustration


Fundamentals Of Differential Equations And Boundary Value Problems
Found in: Page 1
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

Answers without the blur.

Just sign up for free and you're in.


Short Answer

Variation of Parameters. Here is another procedure for solving linear equations that is particularly useful for higher-order linear equations. This method is called variation of parameters. It is based on the idea that just by knowing the form of the solution, we can substitute into the given equation and solve for any unknowns. Here we illustrate the method for first-order equations (see Sections 4.6 and 6.4 for the generalization to higher-order equations).

(a) Show that the general solution to (20)dydx+P(x)y=Q(x) has the formy(x)=Cyh(x)+yp(x), where yh ( 0is a solution to equation (20) when Q(x)=0 ,

C is a constant, andyp(x)=v(x)yh(x) for a suitable function v(x). [Hint: Show that we can take yh=μ-1(x) and then use equation (8).] We can in fact determine the unknown function yhby solving a separable equation. Then direct substitution of vyh in the original equation will give a simple equation that can be solved for v.

Use this procedure to find the general solution to (21) localid="1663920708127" dydx+3xy=x2 , x > 0 by completing the following steps:

(b) Find a nontrivial solution yh to the separable equation (22) localid="1663920724944" dydx+3xy=0 , localid="1663920736626" x>0 .

(c) Assuming (21) has a solution of the formlocalid="1663920777078" yp(x)=v(x)yh(x) , substitute this into equation (21), and simplify to obtain localid="1663920789271" v'(x)=x2yh(x).

d) Now integrate to get localid="1663920800433" vx

(e) Verify that localid="1663920811828" y(x)=Cyh(x)+v(x)yh(x) is a general solution to (21).

  1. Proved
  2. yh=±k.x-3
  3. dvdx=x6
  4. v(x)=x66
  5. Proved
See the step by step solution

Step by Step Solution

Step 1(a): Show that the general solution to (20)dydx+P(x)y=Q(x)   has the formy(x)=Cyh(x)+yp(x) 

Here dydx+P(x)y=Q(x) and yh=μ-1(x)

So the equation is




Rearranging them dyhdx+y(x)P(x)=0

The solution is

localid="1663919860296" width="275" style="max-width: none;" y(x)=1μ(x)μ(x)Q(x)dx+CCyh(x)+yp(x)=1μ(x)μ(x)Q(x)dx+Cyp(x)=v(x)yh(x)=v(x)μ(x)-1=1μ(x)μ(x)Q(x)dx+Cv(x)=μ(x)Q(x)dx

Therefore the general solution is localid="1663920004724" v(x)=μ(x)Q(x)dx

Step 3(b): Find the nontrivial solution

Here dydx+3xy=0

Rearranging them


Raising power e on both sides thenyh=±k.x-3

(k is arbitrary constants)

Therefore the nontrivial solution is yh=±k.x-3.

Step 3(c): Determine the value of  v'(x)

Since yp(x)=v(x)yh(x)and substituteyh=v(x)x-3 then


From part (b) v(x)dyhdx+3xyh=0 then


Where yh=x-3 then


Therefore the value is role="math" localid="1663920371908" dvdx=x6 .

Step 4(d): Find the value v(x)

Integrate the value


Therefore the value is role="math" localid="1663920445772" v(x)=x66 .

Step 5(e): Verify that  y(x)=Cyh(x)+v(x)yh(x) is a general solution to (21)




Therefore, this verifies the result.

Most popular questions for Math Textbooks


Want to see more solutions like these?

Sign up for free to discover our expert answers
Get Started - It’s free

Recommended explanations on Math Textbooks

94% of StudySmarter users get better grades.

Sign up for free
94% of StudySmarter users get better grades.