UNIT 1

                                Ordinary differential equations

Question-1: Find the general solution of   x2 – 4x³

Solution:       rearranging the given equation:

                                             = - =

                         On integrating both sides,

                               y = = 2 In x - + c

Question-2: Find the particular solution of the differential equation  5,                  given the boundary condition y =  1 , when x = 2.

Solution: rearrange the diff. Equation,

                                  =  -

                                 y =   = - + c,    which is the general solution.

     Put the boundary conditions to find c,

                                  1 -    , which gives, c = 1

      Hence the particular solution is,

                                     y = - + 1.

Question-3: :  Find the general solution of   = 3 + 2y

Solution:      here,   = 3 + 2y gives,

                                       ,

Integrating both sides,

                                        ,

By substitution,  u = (3 +2y),

                                             X =  In(3 + 2y) + c.

Question-4: Determine the particular solution of  (y² - 1)3y  given that y =1  when  x = 2.

Solution:  It gives,

When putting the values, y =1 , x = 2 ,

The particular solution will be,

Question-5:   Solve the equation 4xy = y² - 1

Solution:  on separating variables, we get

                                               () dy  =  dx

                                     =  

   Using substitution, u = y² - 1

                                               2In(y² - 1) = In x + c.

Question-6:  Determine the particular solution of =  2 , given that t = 0, when θ = 0

Solution:    =  2  =  2    ,

                                 = 2dt

     dθ =  dt

Now integrating both sides,

  =

The general solution is ,

  + c.

When t = 0 and θ = 0,    c =   

  = 

Question-7:   solve (6x² - y +3)dx + (3y² -x -2)dy = 0

Sol.  First we check exactness,

                                              (3y² -x -2) = -1        

                                             (6x² - y +3) = -1        

                       Hence this is an exact differential equation.

Write the system of equation to find h(x,y),

                         = P(x,y) = (6x² - y +3)

                          = Q(x,y) = (3y² -x -2)

Integrate the first eq. w.r.t. x ,    assume y is constant,

        h(x,y)  =   dx

                     =

                     = 2x³ - xy +3x +g(y)

           Here we have continuous differentiable function g(y) instead of C.

Now,

=  (2x³ - xy +3x +g(y))

                                      -x + g’(x) = 3y² - x – 2

We get,

                         g’(x) = 3y² – 2

Now integrate,

               g(x) = = y³ - 2y

So,

                  h(x,y) = 2x³ -xy +3x +y³ -2y,

The general solution is defined by the following expression,

                       2x³ -xy +3x +y³ -2y = C

Question-8: Determine whether the differential function ydx –xdy = 0 is exact or not.

Solution.  Here the equation is the form of  M(x , y)dx + N(x , y)dy = 0

  But, we will check for exactness,

These are not equal results, so we can say that the given diff. Eq. Is not exact.

Question-9: solve + xy = xy²

Solution :  here , n =2, so me make substitution,

                                                y =

                                                 y’ = 

Now substitute these equations,

                                                    =

                                   Or    z’ – zx = -x  -----------(1)

  this is the linear form of unknown function f(z),

Integrating factor is,

                                     I(x) = =

Multiply (1) by I(x), we get

                                   - x -x

Or                     = -x

      Integrating both sides,

                                        = +c

                                       Z(x) = c

The solution for the original diff. Eq.

                                      y = .

Question-10:  solve .

Solution:  this can be written in differential form,

                      )dx – ydy = 0

 Which can be separated as ,

                                     A(x) = )     , B(y) = -y

Its solution is given by,

                                              - 

                                                 x³  + 2x  -

Solving for y , we obtain the solution in implicit form,

                             y² =   x³  + 4x  + k          with k = -2c

Solving for y,   we get,