# Download PDF by Steven A Leduc: Differential equations By Steven A Leduc

ISBN-10: 0822053209

ISBN-13: 9780822053200

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Extra resources for Differential equations

Example text

Separating the variables and inte grating gives x(v — 1)dv= —(2+3v +v 2 )dx v — 1 ) + 3v + 2 dv dx x (I ) The integral of the left-hand side is evaluated after performing a partial fraction decomposition : v— 1 v 2 +3v+2 v— 1 (v+ 1)(v+2) Therefore, f f( v— 1 v2+3v+2dv 3 -2 1 + v+ 2 v+ ) -2 3 dz v v + 1 +v+2 = -2lnJz~+ 11 +31niv+2 1 =ln(v+ 1) - 22 (v+2)3 1 The right-hand side of (t) immediately integrates to cLr x In+c ' =1nIcx - ' l Therefore, the solution to the separable differential equation (t) i s (v + 1) -2 (v + 2) 3 = cx - ' Now, replacing v by y/x give s y —+ 1 x DIFFERENTIAL EQUATIONS -2 y (— X 3 + 2 =cx - ' FIRST-ORDE R EQUATION S as the general solution of the given differential equation .

CLIFFS QUICK REVIE W 46 FIRST-ORDER EQUATION S (r,9)=(2,n ) ■ Figure 5 ■ Homogeneous Equation s A function equation is said to be homogeneous of degree n if th e f(x, y) f(2x , zy) = z`f(x , y ) holds for all x, y, and z (for which both sides are defined) . Example 18: The function degree 2, sinc e f( zx , zy ) = f(x, y) = x 2 + y2 is homogeneous of (zx ) 2 + (zy ) 2 = z2(x2 + y2) = z 2f(x, y ) • Example 19 : The function f(x, y) = /x8 — 3x 2y h is homogeneous o f degree 4, since f(zv , zy ) = V ( ) 8 - 3 ( ) 2 (zy ) 6 = Vz 8 W - 3x2y6 ) = DIFFERENTIAL EQUATIONS — 3x 2y 6 = z 4f(x, y) ■ 47 FIRST-ORDER EQUATIONS Example 20 : The function f (x, y) = 2x + y is homogeneous of degree 1, sinc e f(zx , zy ) =2 (zx) + (zy ) =z 2x + y) = z' f (x, y) ■ Example 21 : The function f (x, y) = x 3 — y 2 is not homogeneous, since z ,y 2 f (zx, 2y ) _ (zx )3 — (z') 2 = z3x3 — which does not equal z' f (x, y) for any n .

G (x) dx Once this is done, all that is needed to solve the equation is t o integrate both sides . The method for solving separable equation s can therefore be summarized as follows : Separate the variables and integrate . Example 11 : Solve the equation 2y dy (x 2 + 1) dx . _2=4+c c= 4 2 The solution of the IVP is therefore 2)2 3 = 4e-t + x-- + 4 o r 4y 3 42 = e x' + 2x 2 +3 ■ CLIFFS QUICK REVIEW FIRST-ORDER EQUATIONS Example 14: Find all solutions of the differential equatio n (x 2 -1)y 3 dx+x 2 dy=0 .