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Solution Strategies for First Order OrdinaryDiﬀerential Equations MATH 1951: Techniques in Calculus and Diﬀerential Equations Agatha C. Walczak-Typke University of Leeds School of Mathematics  These notes are intended to be an aid for completing Exercise list 4. I give the general strategiesfor solving equations of various kinds, together with general formulas and their derivations. Then,I give model examples of how to solve various equations. Please read this critically, as there is ahigh likelyhood of typos in the text. 1 Equations of the form  M  ( x,y ) dx  +  N  ( x,y ) dy  = 0 1.1 Separation of variables A ﬁrst order DE M  ( x,y ) dx  +  N  ( x,y ) dy  = 0is said to be  separable  if   M  ( x,y ) =  p ( x ) q  ( y ) and  N  ( x,y ) =  r ( x ) s ( y ). Then it can be put into theform f  ( x ) dx  =  g ( y ) dy, where  f  ( x ) =  p ( x ) /r ( x ) and  g ( y ) =  − s ( y ) /q  ( y ). A family of solutions can be obtained by integrationto yield    f  ( x ) dx  =    g ( y ) dy  +  C  1 where  C  1  is an arbitrary constant. The integrals are evaluated, and a function for  y  in terms of   x is found. This general technique is called the method of   separation of variables . Example 1.  Find the general solution of (1 − x ) dy  +  ydx  = 0 . Solution:  First, we separate the variables to yield: dyy  =  dxx − 1 , where  x   = 1 and  y   = 0.Second, we integrate both sides to yield:ln | y | +  C  1  = ln | 1 − x | +  C  2 . Thirdly (this step optional) we solve for  y :ln | y | +  C  1  = ln | 1 − x | +  C  2 e ln | y | + C  1 =  e ln | 1 − x | + C  2 ye C  1 = (1 − x ) e C  2 y  =  C  (1 − x ) . Thus, y  =  C  (1 − x )is the  general solution  .1  1.2 “Homogeneous” equations Sometimes, the best way of solving a DE is a to use a change of variables that will put the DE intoa form whose solution method we know.We now consider a class of DEs that are not directly solvable by separation of variables, but,through a change of variables, can be solved by that method. First a deﬁnition: If a function M  ( x,y ) has the property that  M  ( tx,ty ) =  t n M  ( x,y ) ,  then we say the function  M  ( x,y ) is a  “ho-mogeneous” function of degree  n . The word homogeneous is put in quotes because the sameword is unfortunately used in a diﬀerent context later on.A diﬀerential equation M  ( x,y ) dx  +  N  ( x,y ) dy  = 0is said to be a  “homogeneous” diﬀerential equation  if both functions  M  ( x,y ) and  N  ( x,y ) arehomogeneous functions of (the same) degree  n . In this case, we can represent the functions as M  ( x,y ) =  x n M  (1 ,y/x ) N  ( x,y ) =  x n N  (1 ,y/x )Thus, we can write the diﬀerential equation as dxdy  =  − M  ( x,y ) N  ( x,y ) =  − x n M  (1 ,y/x ) x n N  (1 ,y/x ) =  F  ( y/x ) . Thus, we can substitute y  =  vx ( substitution for “homogeneous” equations )where  v  is a new variable. By diﬀerentiating by x and the application of the product rule, we seethat dydx  =  xdvdx  +  v. Thus, the diﬀerential equation becomes xdvdx  +  v  =  F  ( v ) . Now, the variables can be separated to yield: dvF  ( v ) − v  =  dxx . Example 2.  Solve( x 4 +  y 4 ) dx  + 2 x 3 y dy  = 0 Solution:  First we check that the functions  M  ( x,y ) =  x 4 +  y 4 and  N  ( x,y ) = 2 x 3 y  are both“homogeneous” to the same degree. M  ( tx,ty ) = ( tx ) 4 + ( ty ) 4 =  t 4 M  ( x,y )and N  ( tx,ty ) = 2( tx ) 3 ty  =  t 4 N  ( x,y ) . Thus we see that both functions are “homogeneous” of degree 4.Second, we perform the substitution  y  =  vx . The diﬀerential equation becomes:( x 4 +  x 4 v 4 ) dx  + 2 x 4 v ( v dx  +  xdv ) = 0 . If   x   = 0, then we divide to obtain:(1 + 2 v 2 +  v 4 ) dx  + 2 xv dv  = 0 .  Thirdly, we separate the variables: dxx  + 2 v ( v 2 + 1) 2 dv  = 0 . We integrate to getln | x |− ( v 2 + 1) − 1 =  C  1 . Fourthly, we undo the substitution from earlier using the replacement  v  =  y/x , where  x   = 0 toget:ln | x |−  x 2 x 2 +  y 2  =  C  1 . Note that here it is not so easy to get the solution  y  as a function in terms of   x . Thus, this is oursolution. 1.3 Exact equations If there is a function in two variables  f  ( x,y ) such that in the equation M  ( x,y ) dx  +  N  ( x,y ) dy  = 0it is true that M  ( x,y ) =  ∂f ∂xN  ( x,y ) =  ∂f ∂y then we call the diﬀerential equation an  exact equation . In such a case, the family of solutions of the equation is given by f  ( x,y ) =  C  1 . To check if a given equation of the form  M  ( x,y ) dx  +  N  ( x,y ) dy  = 0 is exact, it is enough toshow that ∂M ∂y  =  ∂N ∂x (conclusive test for exactness) This stems from the fact that for a continuous function  f   in two variables,  ∂  2 f ∂x∂y  =  ∂  2 f ∂y∂x To ﬁnd the function  f  ( x,y ) for a given exact equation, we integrate f  ( x,y ) =    M  ( x,y ) dx  +  g ( y )where the function  g  serves as the “constant” of integration (constant with respect to x). Thefunction  g ( y ) is determined using the function  N  ( x,y ): N  ( x,y ) =  ∂f ∂y  =  ∂ ∂y (    M  ( x,y ) dx  +  g ( y )) =  ∂ ∂y (    M  ( x,y ) dx ) +  g ′ ( y ) . Thus, g ′ ( y ) =  N  ( x,y ) −  ∂ ∂y (    M  ( x,y ) dx )And so, g ( y ) =    N  ( x,y ) dy  −    (  ∂ ∂y (    M  ( x,y ) dx )) dy Finally, we have (a somewhat messy looking) solution f  ( x,y ) =    M  ( x,y ) dx  +    N  ( x,y ) dy  −    (  ∂ ∂y (    M  ( x,y ) dx )) dy  =  C  1  I haven’t put a box around this formula, since you really shouldn’t try to memorize such a thing.The above is just a semi-formal way of showing that  f  ( x,y ) is the integral of   M  ( x,y ) by  x  plus theterms that only involve  y  in the integral of   N  ( x,y ) by  y . Hopefully an example may make thingsclearer: Example 3.  Show that the following diﬀerential equation is exact, and ﬁnd its family of solutions(4 x − 2 y  + 5) dx − (2 x − 2 y ) dy  = 0 Solution:  First, we check that the equation is exact. We identify M  ( x,y ) = (4 x − 2 y  + 5)and N  ( x,y ) =  − (2 x − 2 y ). We calculate that ∂M ∂y  =  − 2 =  ∂N ∂x . Thus, the equation satisﬁes the test for exactness. We now know that there is a function  f  ( x,y )such that ∂f ∂x  = 4 x − 2 y  + 5and ∂f ∂y  =  − (2 x − 2 y ) . To ﬁnd  f  , we second, integrate  M  ( x,y ) by  x  to get: f  ( x,y ) = 2 x 2 − 2 xy  + 5 x  +  g ( y )where  g  is as described above. We will use  N  ( x,y ) to ﬁnd  g .Third, we integrate  N  ( x,y ) by  y  to get: f  ( x,y ) =  y 2 − 2 xy  +  h ( x )Here the function  h ( x ) appears for the same reasons as  g ( y ) above.Fourth, we compare these two results for  f   to see that  g ( y ) =  y 2 (and  h ( x ) = 2 x 2 + 5 x ). Thus,the solution is2 x 2 − 2 xy  + 5 x  +  y 2 =  C  1 . 2 Linear diﬀerential equations  y ′ +  a 0 ( x ) y  =  f  ( x ) First order linear diﬀerential equations have the normal form: y ′ +  a 0 ( x ) y  =  f  ( x ) . If   f  ( x )  ≡  0, then the equation is called  homogeneous . Otherwise, it is called  nonhomogeneous . 2.1 Homogeneous equations We look for solutions of the homogeneous equation y ′ +  a 0 ( x ) y  = 0 . One feature of homogeneous linear DEs in general is that  y  = 0 is a solution. We call this solutionthe  trivial solution . However, we are more interested in  nontrivial  (e.g. non-zero) solutions.Here we can separate variables: dyy  =  − a 0 ( x ) dx, y   = 0 .

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