Lecture 4

     First Order and first-degree differential equation


    A differential equation of the form  M+N dydx=0M+N dydx=0   or  M dx+N dy=0M dx+N dy=0   is called first-order and first-degree differential equation where both  MM and  NN are functions of x and y or constants.

    Types of differential equation of first order and first degree:

            (1) Separation of variables

            (2) Homogeneous equation 

            (3) Equation reducible  to homogeneous

            (4) Exact equation

            (5) Linear equation

            (6) Reducible to linear equation.


    Variable Separable:

If the differential equation    M dx+N dy=0M dx+N dy=0 can be expressed in the form  f(x) dx+ϕ(y) dyf(x) dx+ϕ(y) dy - then it is called variables are separable.
    
    Rule 1: If the differential equation is of the form  f(x) dx+f(y) dy=0f(x) dx+f(y) dy=0 ie function of xx with dxdx and function of yy with dydy, then only integrate we will find the solution
    ie  f(x) dx+f(y) dy=0f(x) dx+f(y) dy=0F(x)+F(y)=cwhen,f(x) dx=F(x)  


   
    
    Question 1: Solve,  dydx+y2+y+1x2+x+1=0  
    Solution: Given equation is  dydx+y2+y+1x2+x+1=0dydx=y2+y+1x2+x+1dyy2+y+1=dxx2+x+1dyy2+y+1+dxx2+x+1=0  
Integrating both sides.  dyy2+y+1+dxx2+x+1=0dy(y+12)2+(32)2+dx(x+12)2+(32)2=0132tan1(y+1232)+132tan1(x+1232)=c23tan1(2y+13)+23tan1(2x+13)=c  



    Rule 2: If any differential equation of the form  f(x) ϕ(y) dx+f(y) ϕ(y) dy=0   then divide the equation by  ϕ(x) ϕ(y) . the equation of the form  f(x)ϕ(x) dx+f(y)ϕ(y) dy=0   ie function of x with dx and function of y with dy. Then only integrate we can find the solution.

   

    
    Question 2: solve,  x1y2 dx+y1x2 dy=0  
    Solution: Given equation is  x1y2 dx+y1x2 dy=0  
Dividing both sides by  1x2 1y2  
  X DX1x2 dx+y dy1y2=0122x dx1x2122y dy1y2=012 21x212 21y2=c     [Since  f(x) dxf(x)=2f(x)]1x2+1y2=c  

   
    

    Question 3: Solve,  sec2x tany dx+sec2y tanx dy=0  
    Solution: Given equation is,  sec2x tany dx+sec2y tanx dy=0  
Dividing both sides by  tanx .tany   we get,
  sec2x dxtanx+sec2y dytany=0  
Now integrating both sides,  sec2x dxtanx+sec2y dytany=0ln(tanx)+ln(tany)=lnc     [Since  f(x)f(x) dx=ln(f(x))]ln(tanx .tany)=lnctanx .tany=c  

    Rule 3: If the differential equation is of the form  dydx=f(ax+by+c)   then put  ax+by+c=z   and simplify. Finally integrating we will get the solution.

   


    Question 4: Solve,  sin1(dydx)=x+y      
    Solution: Given equation is  sin1(dydx)=x+ydydx=sin(x+y)  
We put  x+y=z  
Then  1+dydx=dzdxdydx=dzdx1  
From (4.1) and (4.2) we get,  dzdx1=sinzdzdx=1+sinzdz1+sinz=dx  
Integrating both sides,  dz1+sinz=dx1sinz1sin2z dz=dx1sinzcos2z dz=dxsec2z dz secz .tanz dz=dxtanzsecz=x+ctan(x+y)sec(x+y)=x+c  

   


    Question 5: Solve,  dydx=(4x+y+1)2  
    Solution: Given equation is  dydx=(4x+y+1)2  
we put  4x+y+1=z  
then  4+dydx=dzdxdydx=dzdx4  
From (5.1) and (5.2) we get  dzdx4=z2dzdx=4+z2dz4+z2=dx  
Integrating both sides we get,  12tan1z2=x+c2tan1z2=2x+c        z2=tan(2x+c)4x+y+1=2tan(2x+c)  


   


    Question 6: Solve,  (ex+1)y dy=(y+1)ex dx  
    Solution: The given equation is  (ex+1)y dy=(y+1)ex dxy dyy+1=ex dxex+1(y+11) dyy+1=ex dxex+1(11y+1) dy=d(ex+1)ex+1dydyy+1=dex+1ex+1 
Where the variables have been separated, so the general solution will be obtained by integrating on both sides and adding a constant.
    dydyy+1=d(ex+1)ex+1yln(y+1)=ln(ex+1)+lnc     [wherelnc is the integrating constant]y=ln(y+1)+ln(ex+1)+lncy=ln(y+1)(ex+1).cey=(y+1)(ex+1)c  
Which is the required solution.

   
   

     Question 7: Solve,  y(1+xy) dx+x(1xy) dy=0  
    Solution: The given equation is,  y(1+xy) dx+x(1xy) dy=0y dx+xy2 dx+x dyx2y dy=0x dy+y dx+xy(y dxx dy)=0d(xy)+xy(y dxx dy)=0d(xy)(xy)2+dxxdyy=0   [Dividing (xy)2 on both sides]  
Here the variables have been separated, so the general solution will be obtained by integrating on both sides and adding a constant 
  d(xy)(xy)2+dxxdyy=01xy+lnxlny=clnxlny1xy=clnxy1xy=c  
Which is the required solution. 

   



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