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

1 math(3)

So Solu lutio tion n of of fi firs rst t or orde der r ode ode 2 math(3)

Preface The first order differential equations may appear in one of the following forms: dy dx  f x y ( , )    dy   F x y dx , , 0     M x y dx ( , ) N x y dy ( , ) 0 3 math(3)

Methods of Solution of First Order ODE (1) Separable Differential Equations: dy g x ( )   dy      dx  h y dy ( ) g x dx ( ) c f x y ( , ) dx h y ( )            a x b y dx c x d y dy ( ) 0 M x y dx ( , ) N x y dy ( , ) 0 4 math(3)

Separable Differential Equations A separable differential equation can be expressed as the product of a function of x and a function of y . dy       dx   h y  g x h y 0 Example: Multiply both sides by dx and divide dy dx  2 2 xy both sides by y 2 to separate the variables. (Assume y 2 is never zero.) dy y  2 x dx 2   2 y dy 2 x dx math(3) 5

Separable Differential Equations A separable differential equation can be expressed as the product of a function of x and a function of y . dy       dx   h y  g x h y 0 Example:     2 y dy 2 x dx dy dx  2 2 xy Combined      1 2 y C x C constants of 1 2 integration dy 1 y     2 2 x dx x C 2 y 1 1     y y   2  y dy 2 x dx  2 2 x C x C math(3) 6 

Initial conditions • In many physical problems we need to find the particular solution that satisfies a condition of the form y(x 0 )=y 0 . 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 . math(3) 7

Example:   dy dx   2 2 x 2 x 1 y e Separable differential equation 1  2 x dy 2 x e dx  2 1 y  1 2   u x  2 x dy 2 x e dx  2  1 y du 2 x dx 1    u dy e du  2 1 y     1 u tan y C e C 1 2    2  1 x tan y C e C 1 2   2  math(3) 8 1 x tan y e C  Combined constants of integration

Example (cont.):   dy dx   2 2 x 2 x 1 y e   2  1 x tan y e C We now have y as an implicit function of x .       2  1 x We can find y as an explicit tan tan y tan e C function of x by taking the tangent   of both sides.  2  x y tan e C math(3) 9 

Example dy       Solve the IVP x y 1 , e y 0 1 dx Solution dy  dy    x y 1  x y 1 e e e dx dx     x   1 y    x 1 y e e c e dx e dy c         x 0 y 1 1 1 c c 2 e     x 1 y e 2 0 10 math(3)

Example   dy      2 2 x y 2 x xy 2 y x 2 Solve the DE dx Solution   dy 2  2     x y 2 x xy 2 y x 2 dx 2 (      x y 2) dy y x ( 2) ( x 2) dx 2 (     x y 2) dy ( x 2)( y 1) dx   1 1 2   y 2 x 2      1 1) dy dx c dy dx  2  ( y x x 2 y 1 x 2      y ln( y 1) ln x c x 11 math(3)

(2) DE ’ s Reducible to Separable This type takes the form: dy dx    f ax ( by c )    u ax by c Put this transform the DE into separable one. 12 math(3)

Example dy dx    2 Solve the DE (2 x y 7) Solution    l t e u 2 x y 7 differentiate w.r.t. x , we get: du   u   dx c  2 du dy 2   2     dx dx 1 tan u 1    x c du   dx   2 2 2 2 u      1 2 x y 7   1   tan x c du   dx   2 2 2 2 u 13 math(3)

Example dy Solve the DE dx    sin( x y 1) Solution    l et u x y 1 du   u   differentiate w.r.t. x , we get: dx c  1 sin du dy   1  1 sin u du     dx dx dx c  2 1 sin u du dx   1 sin u       2 2 sec u (cos ) u sin u du dx c du dx     1   1 sinu tan u (cos u ) x c 14 math(3)

(3) Homogeneous DE ’ s   dy y    f This type takes the form:   dx x The right hand side is a homogeneous function of degree zero. y  u x This transform the DE into separable one. 15 math(3)

Example y y dy dy y dx     x x Solve the DE: x y x e e dx x Solution y    l et u y u x x du dx  u x e differentiate w.r.t. x , we get: 1 dy du        u e du dx c u x x dx dx     du u    u e ln x c u x u e dx 16 math(3)

Example y  1  dy dy x y x   Solve the DE: y  dx  dx x y 1 x Solution y     du 1 u l et u y u x   x u x  dx 1 u differentiate w.r.t. x , we get:    2 du 1 u u u  x dx  dy du 1 u   u x  1 u du 1 dx dx     dx c   2 1 u x du 1 u   u x dx   1       1 u 1 2 tan u ln 1 u ln x c 2 17 math(3)

example The slope m of a curve is 0 where the curve 𝟐 + 𝒛 𝟑 . Find m as a 𝒆𝒛 crosses the y-axis, and 𝒆𝒚 = function of x. SOLUTION 𝑒𝑧 1+𝑧 2 = 𝑒𝑦 Integrate both sides 𝑒𝑧 𝒕𝒋𝒐𝒊 −𝟐 (y)=x+c 1+𝑧 2 = 𝒆𝒚 + 𝒅 y=0 at x=0 then c=0 and y=sinh(x) 18 math(3)

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