1 math(3) So Solu lutio tion n of of fi firs rst t or - - PowerPoint PPT Presentation

math(3) 1

So Solu lutio tion n of

• f

fi firs rst t or

• rde

der r

• de
• de

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Preface

The first order differential equations may appear in one of the following forms:

( , ) dy f x y dx 

, , dy F x y dx       

( , ) ( , ) M x y dx N x y dy  

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Methods of Solution of First Order ODE

( , ) dy f x y dx 

( ) ( ) dy g x dx h y 



( , ) ( , ) M x y dx N x y dy  

(1) Separable Differential Equations:



     

( ) a x b y dx c x d y dy  



( ) ( ) h y dy g x dx c  

 

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Separable Differential Equations

A separable differential equation can be expressed as the product of a function of x and a function of y.

   

dy g x h y dx  

Example:

2

2 dy xy dx 

Multiply both sides by dx and divide both sides by y2 to separate the

• variables. (Assume y2 is never zero.)

2

2 dy x dx y 

2

2 y dy x dx





 

h y 

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A separable differential equation can be expressed as the product of a function of x and a function of y.

   

dy g x h y dx  

Example:

2

2 dy xy dx 

2

2 dy x dx y 

2

2 y dy x dx





2

2 y dy x dx





 

1 2 1 2

y C x C



   

2

1 x C y   

2

1 y x C   

2

1 y x C   

 

h y 



Combined constants of integration

Separable Differential Equations

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Initial conditions

• In many physical problems we need to find the

particular solution that satisfies a condition of the form y(x0)=y0. This is called an initial condition, and the problem of finding a solution of the differential equation that satisfies the initial condition is called an initial-value problem.

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Example:

 

2

2

2 1

x

dy x y e dx  

2

2

1 2 1

x

dy x e dx y  

Separable differential equation

2

2

1 2 1

x

dy x e dx y  

 

2

u x  2 du x dx 

2

1 1

u

dy e du y  

 

1 1 2

tan

u

y C e C



  

2

1 1 2

tan

x

y C e C



  

2

1

tan

x

y e C



 

Combined constants of integration



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Example (cont.):

 

2

2

2 1

x

dy x y e dx  

2

1

tan

x

y e C



 

We now have y as an implicit function of x. We can find y as an explicit function of x by taking the tangent

• f both sides.

 

 

2

1

tan tan tan

x

y e C



 

 

2

tan

x

y e C  



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Example

 

1,

1

x y

dy e y dx

 

  Solve the IVP

1 x y

dy e dx

 



1 x y

dy e e dx





Solution

1 x y

e dx e dy c



 

 

1 x y

e e c



  

1 x y   

2 1 1      c c

1

2

x y

e e    

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Example

 

2 2

2 2 2 dy x y x xy y x dx      Solve the DE

Solution

 

2 2

2 2 2 dy x y x xy y x dx     

2(

2) ( 2) ( 2) x y dy y x x dx     

2(

2) ( 2)( 1) x y dy x y dx    

2

2 2 1 y x dy dx y x    

2

1 1 2 1 ( 1)dy dx c y x x     

 

2 ln( 1) ln y y x c x     

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(2) DE’s Reducible to Separable

This type takes the form:

u ax by c   

Put

( ) dy f ax by c dx   

this transform the DE into separable one.

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Example

2

(2 7) dy x y dx    Solve the DE

Solution

2 7 u y l t x e   

2 du dy dx dx  

2

2 du u dx  

2

2 du u dx  

2

2 du dx c u   

 

1

1 tan 2 2 u x c

 

      

differentiate w.r.t. x, we get:

1

1 2 7 tan 2 2 x y x c



         

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Example

sin( 1) dy x y dx    Solve the DE

Solution

1 et u y l x   

1 du dy dx dx  

1 sin du u dx  

1 du sinu dx  

1 sin du dx c u   

 

2

1 sin 1 sin u du dx c u    

 

differentiate w.r.t. x, we get:

2 2

sec (cos ) sin u u u du dx c



  

 

c x u u   

1

) (cos tan

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(3) Homogeneous DE’s

This type takes the form:

y u x 

The right hand side is a homogeneous function of degree zero.

dy y f dx x       

This transform the DE into separable one.

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Example

y x

dy x y x e dx   Solve the DE:

Solution

y et u x l u y x    dy du u x dx dx  

u

du u x u e dx   

u

du x e dx 

1

u

e du dx c x



 

 

ln

u

e x c



  

differentiate w.r.t. x, we get:

y x

dy y e dx x  

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Example

dy x y dx x y    Solve the DE:

Solution

y et u x l u y x    dy du u x dx dx  

1 1 du u u x dx u    

1 1 du u x u dx u    

differentiate w.r.t. x, we get:

1 1 y dy x y dx x   

2

1 1 du u u u x dx u     

2

1 1 1 u du dx c u x    

 

 

c x u u    



ln 1 ln 2 1 tan

2 1

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example

The slope m of a curve is 0 where the curve crosses the y-axis, and

𝒆𝒛 𝒆𝒚 =

𝟐 + 𝒛𝟑 . Find m as a function of x. SOLUTION

𝑒𝑧 1+𝑧2 = 𝑒𝑦

Integrate both sides

𝑒𝑧 1+𝑧2 = 𝒆𝒚 + 𝒅

𝒕𝒋𝒐𝒊−𝟐(y)=x+c y=0 at x=0 then c=0 and y=sinh(x)

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