Differential Equations - 19 - Mathematics MCQ

Proof: The given differential equation is
(y - px) (p - i) = p ...... (i)
  ......(ii)
Differentiating (ii) w.r. t. x, we get



⇒ p = c     .......(iii)
Eliminating p from equations (ii) and (iii),

Proof : The given differential equation is 


Substituting from (ii) and (iii) in (i), we get


or v = up + p²  ......(iv)

Thus equation (iv) is in Ciairaut's form.

The differential equation in statement (d) can not be reduced to Ciairaut’s form by the transformation
x² = u and y² = v    ....(i)
Proof : The transform ation (i) yields
  .....(ii)
Substitute in the given differential equation from (i) and (ii) and find that the transformed equation is not in Ciairaut’s form.

Hint : The transform aticn xy = v and y = u
implies that

The given equation will be reduced to
v = up + p²
which is in the Clairaut's form.

If the given differential equation is in Ciairaut's form, then any one of the p and c discriminants can be used for finding the singular solution of the differential equation. 
Remark : The equation in Clairaut’s form is
y = xp +f(p)    ...(i)
Its general solution is given by
y = cx + f(d)    ...(ii)
Differentiating (ii) partially w.r. t. c, 
0 = x + f'(c)       ...(iii)
The singular solution which is the envelope of (ii) can now be obtained by eliminating c from (ii) and (iii).
Alternately : Differentiate (i) partially w.r. t. p,
0 = x + f'(P)    ...(iv)
The singular solution can now be obtained by eliminating p from (i) and (iv).

Proof : The given differential equation is P + = 1 ...(i)
  ...(iii)


⇒ y = sin (x + c) ...(iii)
Differentiating (iii) w.r.t. c, we get 0 = cos (x + c)
⇒ x + c - π/2 ...(iv)
Substituting for x + c from (iv) in (iii), one singular solution is given by


This gives another singular solution as y = -1.
∴ There are 2 singular solutions
Note : complete case - 2.

Obtain all the four singular solutions.

Proof : The given differential equation is


  ......(i)

  ...(ii)
Differentiating (ii)partially with respect to c, we get

Substituting for c from (iii) in (i), one singular solution is given by

or x = 0
Similarly, if we rewrite (i) as

then another singular solution will be y = 0 
∴ x = 0 arid y = 0
are both singular solutions.

Euler's method provided the solution of all I he first order differential equations of the form

Euler’s method.
This method provides the solution of the differential equations of the form

in the form of a set of tabulated values. This method is very slow and to obtain reasonable accuracy with Euler's method, h should be taken very small. This method provid es an approximate solution of equation (i) in general.

Proof : Integrating the differential equation

If the coefficients a₀, a₁ and a₂ are constants in the differential eauation.

then it will be adifferential equation with constant coefficients.
Remark : There is no conditon on f(x) and it may be a constant (zero or non-zero) or a function of x.