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Q 2.6-36E

Fundamentals Of Differential Equations And Boundary Value Problems
Found in: Page 77
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

Question: In Problems 33–40, solve the equation given in: Problem 4.

The solution of the given equation is -23tan-13(t-1)(x+3)+ln(3t-12+(x+3)2=C

See the step by step solution

Step by Step Solution

Step 1: Concept and definition

Homogeneous differential equation is the equation of the form fx,ydy=gx,ydx , where the degree of fx,y and gx,y is same.

Step 2: Analyzing the given statement

One has to solve the equation given in problem 4, i.e.


Step 3: Transform the equation (1) in homogeneous equation in u and v

Comparing (1) with the equation,

a1x+b1y+c1dx+a2x+b2y+c2dy=0As a1b2a2b1

Here, take x as the dependent variable and t as the independent variable.

Let t=u+h and x=v+k , where h and k satisfy,


Solving (2) and (3),

role="math" localid="1663929399330" h=1andk=-3

Now as t=u+h and x=v+k

Therefore, dt=du and dx=dv

Substituting these values of x, t, dx, dt in equation (1),


which is the required homogeneous equation in u and v.

Step 4: Find solution of equation (4)

Let vu=z in equation (4)

Differentiating both sides,


Substitute this value of localid="1663929574069" dvdu in equation (4),


Separating the variables,


Integrating both sides,


Now, solve the integration part then


Now, equation (4) becomes,



Hence, the solution of given equation is -23tan-13(t-1)(x+3)+ln(3t-12+(x+3)2=C.

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