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Buenavista Community College

This document discusses reducing higher order differential equations to first order equations. Specifically, it addresses two cases: 1) When the dependent variable is absent, the equation can be reduced by substituting dy/dx for p and d^2y/dx^2 for dp/dx. 2) When the independent variable is absent, the equation can be reduced by substituting dy/dx for p and d^2y/dx^2 for p*dp/dy. Several examples of each type are worked out. The document concludes with an application section that provides additional practice problems, including finding general and particular solutions for specific differential equations.

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Oliver Estoce
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75 views3 pages

Buenavista Community College

This document discusses reducing higher order differential equations to first order equations. Specifically, it addresses two cases: 1) When the dependent variable is absent, the equation can be reduced by substituting dy/dx for p and d^2y/dx^2 for dp/dx. 2) When the independent variable is absent, the equation can be reduced by substituting dy/dx for p and d^2y/dx^2 for p*dp/dy. Several examples of each type are worked out. The document concludes with an application section that provides additional practice problems, including finding general and particular solutions for specific differential equations.

Uploaded by

Oliver Estoce
Copyright
© © All Rights Reserved
We take content rights seriously. If you suspect this is your content, claim it here.
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Download as DOCX, PDF, TXT or read online on Scribd
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BUENAVISTA COMMUNITY COLLEGE

Cangawa, Buenavista, Bohol


Telefax: (038)5139169/Tel.: 513-9179

Second Order Equations Reducible to First Order


I. Overview:
In this module, we shall discuss the certain types of differential equations of higher
order that can be solve by reducing to lower order DE.

II. Specific Objectives: At the end of the lesson, the students are expected to:
a)reduce higher order to lower order DE;
b) solve the general solution of a higher order DE;
c) find the particular solution with a given condition of a DE.
III. Learning Session:
Certain types of differential equations of higher order can be solved reducing
them to equivalent equations of lower order. We shall discuss two such
reductions.
A. Dependent variable absent. If the dependent variable y is missing, a second
order equation can be reduced to a first order equation by the substitution
dy d2 y d p
= p, = .
dx dx 2 dx

If this first order equation can be solved for p in terms of x, then an additional
integration will give the required solution.
Example 1. Solve y’y” = 1.
dp
Solution: Placing y’ = p and y” = , we obtain
dx
dp
p = 1; hence p2 = 2x + c1.
dx
Solving for p and integrating, we find
dy 1
p= = ±√ 2 x +c 1 ; therefore y=± (2x + c1)3/2 + c2.
dx 3

Example 2. Find the general solution of the DE xy” – y’ = 0.

Math 401 - Differential Equation (Module 5) 1


dp
Solution: Placing y’ = p and y” = , we obtain
dx
dp
x –p=0
dx
Separating variables and integrating
dp dx
p
= x
dy
ln p = ln x + c1, hence, p = c1x =
dx
Therefore, y = c1x2 + c2.

B. Independent variable absent. If the independent variable x is missing, a


second order equation can be reduced to a first order equation by the
substitution

dy d 2 y dp dy dp
= p, 2 = ∙ =p .
dx dx dy dx dy

Example 1. Solve y” = yy’.


dp
Solution: Placing y’ = p and y” = p , we obtain
dy
dp
p = yp; hence p = ½ y2 + C.
dy
Separating variables, we have
dy
dx = 1 2 .
y +C
2
Therefore, on integrating, we find
x = 2c1 Tan-1 c1y + c2,
where c1 = (2C)-1/2 and C>0.

Note: If the constant C is negative, the solution of the above differential equation has
the form

1 y−c 1
x= ln + c2, where c1 = √ −2 C .
c 1 y +c 2

Math 401 - Differential Equation (Module 5) 2


IV. Application of Learning:

A. Find the general solution of each of the following differential equations.


1) y’y” = x. 4) y” = y’2 – y’.

2) (1 + x2)y” = 1 + y’2. 5) xy” + y’ = 4x.

3) y” + y’2 = 1. 6) y” + 2yy’ = 2y’

B. For the following differential equations find the particular solution which
satisfies the given conditions.
1) y’y” = x; y = 0 and y’ = 0, when x = 0.

2) x2y” + xy’ = 1; y = 0 and y’ = ½ , when x = 1.

3) yy” + y’2 = 1; y = 2 when x = 0 or 1.

4) Find the equation of the parabola whose differential equation


is yy” + y’2 = 0 and which passes through the points (-1,0) and
(3,4).

Note: Write your answer in a yellow paper and pass it during our meeting. This will serve as your quiz.

Math 401 - Differential Equation (Module 5) 3

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