[<< wikibooks] Ordinary Differential Equations/Separable 4
== Existence problems ==
1) f(x,y) has no discontinuities, so a solution exists.  
  
    
      
        
          
            
              ∂
              
                f
              
            
            
              ∂
              
                y
              
            
          
        
      
    
    {\displaystyle {\frac {\partial {f}}{\partial {y}}}}
   has no discontinuities, so the solution is unique.
2) f(x,y) is not defined for the point (-1,10) because ln(x) is not defined.  So no solution exists.
3) f(x,y) has discontinuities at y=1 and -1, but not at 0 so a solution exists.  
  
    
      
        
          
            
              ∂
              
                f
              
            
            
              ∂
              
                y
              
            
          
        
      
    
    {\displaystyle {\frac {\partial {f}}{\partial {y}}}}
   has no discontinuities at (0,16) so the solution is unique.
4) f(x,y) has discontinuities at y<0, but not at 1 so a solution exists.  
  
    
      
        
          
            
              ∂
              
                f
              
            
            
              ∂
              
                y
              
            
          
        
      
    
    {\displaystyle {\frac {\partial {f}}{\partial {y}}}}
   is discontinuous at 1, so the solution is not unique
5) f(x,y) has discontinuities at -3 and -4, but not at 0 so a solution exists.  
  
    
      
        
          
            
              ∂
              
                f
              
            
            
              ∂
              
                y
              
            
          
        
      
    
    {\displaystyle {\frac {\partial {f}}{\partial {y}}}}
   has no discontinuities at (5,9) so the solution is unique.
6) f(x,y) has a discontinuity at x=5, so no solution exists.


== Separable equations ==
7) 
  
    
      
        
          y
          ′
        
        =
        
          y
          
            3
          
        
        s
        e
        
          c
          
            2
          
        
        (
        x
        )
      
    
    {\displaystyle y'=y^{3}sec^{2}(x)}
  

  
    
      
        
          
            
              d
              y
            
            
              y
              
                3
              
            
          
        
        =
        s
        e
        
          c
          
            2
          
        
        (
        x
        )
        d
        x
      
    
    {\displaystyle {\frac {dy}{y^{3}}}=sec^{2}(x)dx}
  

  
    
      
        ∫
        
          
            
              d
              y
            
            
              y
              
                3
              
            
          
        
        =
        ∫
        s
        e
        
          c
          
            2
          
        
        (
        x
        )
        d
        x
      
    
    {\displaystyle \int {\frac {dy}{y^{3}}}=\int sec^{2}(x)dx}
  

  
    
      
        −
        
          
            1
            
              2
              
                y
                
                  2
                
              
            
          
        
        =
        t
        a
        n
        (
        x
        )
        +
        C
      
    
    {\displaystyle -{\frac {1}{2y^{2}}}=tan(x)+C}
  

  
    
      
        y
        =
        −
        
          
            1
            
              (
              2
              t
              a
              n
              (
              x
              )
              +
              C
              )
            
          
        
      
    
    {\displaystyle y=-{\frac {1}{\sqrt {(2tan(x)+C)}}}}
  

8) 
  
    
      
        
          y
          ′
        
        =
        
          
            
              5
              
                y
                
                  2
                
              
              +
              6
            
            y
          
        
      
    
    {\displaystyle y'={\frac {5y^{2}+6}{y}}}
  

  
    
      
        
          
            
              y
              d
              y
            
            
              5
              
                y
                
                  2
                
              
              +
              6
            
          
        
        =
        d
        x
      
    
    {\displaystyle {\frac {ydy}{5y^{2}+6}}=dx}
  

  
    
      
        ∫
        
          
            
              y
              d
              y
            
            
              5
              
                y
                
                  2
                
              
              +
              6
            
          
        
        =
        ∫
        d
        x
      
    
    {\displaystyle \int {\frac {ydy}{5y^{2}+6}}=\int dx}
  

  
    
      
        
          
            1
            10
          
        
        l
        n
        (
        5
        
          y
          
            2
          
        
        +
        6
        )
        =
        x
        +
        C
      
    
    {\displaystyle {\frac {1}{10}}ln(5y^{2}+6)=x+C}
  

  
    
      
        y
        =
        ±
        
          
            C
            
              e
              
                10
                x
              
            
            −
            
              
                6
                5
              
            
          
        
      
    
    {\displaystyle y=\pm {\sqrt {Ce^{10x}-{\frac {6}{5}}}}}
  

9) 
  
    
      
        
          y
          ′
        
        =
        
          x
          
            3
          
        
        
          /
        
        
          y
          
            3
          
        
      
    
    {\displaystyle y'=x^{3}/y^{3}}
  

  
    
      
        
          y
          
            3
          
        
        d
        y
        =
        
          x
          
            3
          
        
        d
        x
      
    
    {\displaystyle y^{3}dy=x^{3}dx}
  

  
    
      
        ∫
        
          y
          
            3
          
        
        d
        y
        =
        ∫
        
          x
          
            3
          
        
        d
        x
      
    
    {\displaystyle \int y^{3}dy=\int x^{3}dx}
  

  
    
      
        
          
            1
            4
          
        
        
          y
          
            4
          
        
        =
        
          
            1
            4
          
        
        
          x
          
            4
          
        
        +
        C
      
    
    {\displaystyle {\frac {1}{4}}y^{4}={\frac {1}{4}}x^{4}+C}
  

  
    
      
        y
        =
        (
        
          x
          
            4
          
        
        +
        C
        
          )
          
            
              1
              4
            
          
        
      
    
    {\displaystyle y=(x^{4}+C)^{\frac {1}{4}}}
  

10) 
  
    
      
        
          y
          ′
        
        =
        
          x
          
            2
          
        
        +
        3
        x
        −
        9
      
    
    {\displaystyle y'=x^{2}+3x-9}
  

  
    
      
        d
        y
        =
        (
        
          x
          
            2
          
        
        +
        3
        x
        −
        9
        )
        d
        x
      
    
    {\displaystyle dy=(x^{2}+3x-9)dx}
  

  
    
      
        ∫
        d
        y
        =
        ∫
        (
        
          x
          
            2
          
        
        +
        3
        x
        −
        9
        )
        d
        x
      
    
    {\displaystyle \int dy=\int (x^{2}+3x-9)dx}
  

  
    
      
        y
        =
        
          
            1
            3
          
        
        
          x
          
            3
          
        
        +
        
          
            3
            2
          
        
        
          x
          
            2
          
        
        −
        9
        x
        +
        C
      
    
    {\displaystyle y={\frac {1}{3}}x^{3}+{\frac {3}{2}}x^{2}-9x+C}
  

11) 
  
    
      
        
          y
          ′
        
        =
        c
        o
        s
        (
        y
        )
        
          /
        
        s
        i
        n
        (
        y
        )
      
    
    {\displaystyle y'=cos(y)/sin(y)}
  

  
    
      
        
          
            
              s
              i
              n
              (
              y
              )
              d
              y
            
            
              c
              o
              s
              (
              y
              )
            
          
        
        =
        d
        x
      
    
    {\displaystyle {\frac {sin(y)dy}{cos(y)}}=dx}
  

  
    
      
        ∫
        
          
            
              s
              i
              n
              (
              y
              )
              d
              y
            
            
              c
              o
              s
              (
              y
              )
            
          
        
        =
        ∫
        d
        x
      
    
    {\displaystyle \int {\frac {sin(y)dy}{cos(y)}}=\int dx}
  

  
    
      
        −
        l
        n
        (
        c
        o
        s
        (
        y
        )
        )
        =
        x
        +
        C
      
    
    {\displaystyle -ln(cos(y))=x+C}
  

  
    
      
        y
        =
        a
        r
        c
        c
        o
        s
        (
        C
        
          e
          
            x
          
        
        )
      
    
    {\displaystyle y=arccos(Ce^{x})}
  

12) 
  
    
      
        
          y
          ′
        
        =
        
          
            
              c
              o
              s
              (
              x
              )
            
            
              s
              i
              n
              (
              y
              )
            
          
        
      
    
    {\displaystyle y'={\frac {cos(x)}{sin(y)}}}
  

  
    
      
        s
        i
        n
        (
        y
        )
        d
        y
        =
        c
        o
        s
        (
        x
        )
        d
        x
      
    
    {\displaystyle sin(y)dy=cos(x)dx}
  

  
    
      
        ∫
        s
        i
        n
        (
        y
        )
        d
        y
        =
        ∫
        c
        o
        s
        (
        x
        )
        d
        x
      
    
    {\displaystyle \int sin(y)dy=\int cos(x)dx}
  

  
    
      
        −
        c
        o
        s
        (
        y
        )
        =
        s
        i
        n
        (
        x
        )
        +
        C
      
    
    {\displaystyle -cos(y)=sin(x)+C}
  

  
    
      
        y
        =
        a
        r
        c
        c
        o
        s
        (
        −
        s
        i
        n
        (
        x
        )
        +
        C
        )
      
    
    {\displaystyle y=arccos(-sin(x)+C)}
  


== Initial value problems ==
13) 
  
    
      
        
          y
          ′
        
        =
        c
        o
        s
        (
        x
        )
        +
        s
        i
        n
        (
        x
        )
        ,
        y
        (
        0
        )
        =
        1
      
    
    {\displaystyle y'=cos(x)+sin(x),y(0)=1}
  

  
    
      
        d
        y
        =
        (
        c
        o
        s
        (
        x
        )
        +
        s
        i
        n
        (
        x
        )
        )
        d
        x
      
    
    {\displaystyle dy=(cos(x)+sin(x))dx}
  

  
    
      
        ∫
        d
        y
        =
        ∫
        (
        c
        o
        s
        (
        x
        )
        +
        s
        i
        n
        (
        x
        )
        )
        d
        x
      
    
    {\displaystyle \int dy=\int (cos(x)+sin(x))dx}
  

  
    
      
        y
        =
        s
        i
        n
        (
        x
        )
        −
        c
        o
        s
        (
        x
        )
        +
        C
      
    
    {\displaystyle y=sin(x)-cos(x)+C}
  

  
    
      
        1
        =
        s
        i
        n
        (
        0
        )
        −
        c
        o
        s
        (
        0
        )
        +
        C
        =
        0
        −
        1
        +
        C
        =
        C
        −
        1
      
    
    {\displaystyle 1=sin(0)-cos(0)+C=0-1+C=C-1}
  

  
    
      
        C
        =
        2
      
    
    {\displaystyle C=2}
  

  
    
      
        y
        =
        s
        i
        n
        (
        x
        )
        −
        c
        o
        s
        (
        x
        )
        +
        2
      
    
    {\displaystyle y=sin(x)-cos(x)+2}
  

14) 
  
    
      
        
          y
          ′
        
        =
        7
        
          y
          
            2
          
        
        ,
        y
        (
        5
        )
        =
        9
      
    
    {\displaystyle y'=7y^{2},y(5)=9}
  

  
    
      
        
          
            
              d
              y
            
            
              y
              
                2
              
            
          
        
        =
        7
        d
        x
      
    
    {\displaystyle {\frac {dy}{y^{2}}}=7dx}
  

  
    
      
        ∫
        
          
            
              d
              y
            
            
              y
              
                2
              
            
          
        
        =
        ∫
        7
        d
        x
      
    
    {\displaystyle \int {\frac {dy}{y^{2}}}=\int 7dx}
  

  
    
      
        −
        
          
            1
            y
          
        
        =
        7
        x
        +
        C
      
    
    {\displaystyle -{\frac {1}{y}}=7x+C}
  

  
    
      
        y
        =
        
          
            1
            
              −
              7
              x
              +
              C
            
          
        
      
    
    {\displaystyle y={\frac {1}{-7x+C}}}
  

  
    
      
        9
        =
        
          
            1
            
              −
              7
              ∗
              5
              +
              C
            
          
        
      
    
    {\displaystyle 9={\frac {1}{-7*5+C}}}
  

  
    
      
        C
        =
        
          
            316
            9
          
        
      
    
    {\displaystyle C={\frac {316}{9}}}
  

  
    
      
        y
        =
        
          
            1
            
              −
              7
              x
              +
              
                
                  316
                  9
                
              
            
          
        
      
    
    {\displaystyle y={\frac {1}{-7x+{\frac {316}{9}}}}}