Differential Equations - 16 - Mathematics MCQ

I Proof: The family of circles of radius a is given by
(x - h)² + (y - k)² = a²,    .......(i)
where h and k are the parameters to be eliminated. Since there are two arbitrary constants, therefore equation (i) is to be differentiated twice and the order of the differential equation must be 2.
Now differentiating (i) twice, we get

and

Eliminating k and h from equations (i), (ii) and (iii), the required differential equation of the circles of radius a is given by

Definition : A differential equation is said to be linear if.
i. the dependent variable and each of its derivatives occurs to the first degree only and
ii. no products of dependent variable and or its derivatives occur.

Remark : Observe that the dependent variable y and its derivatives

occur to the first degree and there are no products of them in statement (d).
The product of y and dy/dx occurs in (a), (b) and (c)

The differential equations in (a), (b) and (c) are of first degree but none of them is linear.

(i)(a) is not correct because a differential equation of first degree may not be linear. For example, the differential equation.

is of first degree but non-linear, 
(ii) (b) is not correct because a differential equation of first, degree is not necessarily of first order. The differential equation (i) is of first, degree but of second order.
(iii) (d) is not correct because a differential equation of first degree may be of first order and may not be linear.

Any differential equation which is not of first degree can not be linear.

The differential equation

is of first order and first degree. However, it may or may not be linear.
M(y, y)dx + N(x, y)dy = 0
can be rewritten as

and therefore is of first degree and first order. It also may or may not be linear.

The necessary and sufficient condition for the differential equation
M(x, y)dx + N(x, y)dy = 0 to be exact is that

Proof: Condition


Remark : An exact differential equation can be written as du = 0. where u is some function of x and y.

Proof : The given differential equation is
P(x)dx+Q(y)dy = 0 ...(i)
Comparing it with
Mdx + Ndy = 0 ...(ii) 
∴ M = P(x) and N = Q(y)

is always satisfied.
⇒ D.E. (i) is always exact.

Proof : The given differential equation is
P(y)dx+Q(x)dy = 0 ...(i)
Comparing (i) with
Mdx + Ndy= 0 ...(ii)
M = P(y) and N = Q(x)
Differential equation (i) will be exact if and only if

    ...(iii)

Consider the differential equation in (a),
(y² + x) dx + (y³ + x)dy = 0 ... (i)

so that DE (i) is not exact.
(ii) The differential equation in (b) can be rewritten as
(x + y - 1) dx - (x + y + 1 ) = 0  .....(ii)


so that DE (ii) is not exact
(iii) The differential equation in (c) is
2xy dx + (x + y²) dy = 0  .....(iii)


so that DE (iii) is not exact.
Thus none of the differential equations in (a), (b) and (c) is exact.

The differential equation in (a) is exact.
Remark : Verify yourself that (a) is exact but (b), (c) or fd) are not exact.

The given differential equation can be written as

The given differential equation can be rewritten as

The expression in (a) is not exact. Hence (a) is the correct answer.
Remark : Note that expression in (a)

Comments about Integrating Factor 
Definition : Suppose that tne differential equation. M dx + N dy = 0    ...(i)
is not exact but the differential equation

is exact, where μ =  F(x, y) for a suitably chosen function F. Then μ is called an integrating factor of the differential equation (i).
Ex. : The differential equation
(3ty + 4xy²) dx + (2x + 3x²y) dy = 0    ...(ii)
is not exact. But if we choose
μ = x² y.
then the differential equation

becomes exact.
∴ μ = x² y is an integrating factor of (ii).
Rules for finding an Integrating Factor :
Rule : 1: If Mx ± Ny ≠ 0. and is homogeneous in x and y then the integrating factor is

Ex. : Consider the differential equation
(x²y - 2xy²)dx- (x³ - 3x²y)dy = 0    ...(iii)
verify that
(i) equation (iii) is noi exact and (ii)

Intergrating factor is given by

Hence the equation

should be exact. Now equation (iv) can be written as 


Rule : 2. If 

is a function of x alone, say f(x), then the integrating factor μ is given by

Rule : 3. If

is a function of y alone, say φ(y), then the integrating factor μ is given by

To solve a differential equation of the form dx/P(x, y) = dy/Q(x, y), one method is to multiply both sides of the equation by an integrating factor, which is a function of x and y that can be chosen to make the left-hand side of the equation an exact differential.

An exact differential is a differential that can be expressed as the derivative of some function, i.e. dF = P(x, y) dx + Q(x, y) dy. If a differential equation is exact, then it can be easily solved by finding the function F such that its derivative is equal to the left-hand side of the equation.

In this case, the given differential equation is not exact, because the left-hand side is not the derivative of any function. However, we can make it exact by multiplying both sides by an integrating factor.

To find the integrating factor, we need to find a function I(x, y) such that dI/I = P(x, y) dx + Q(x, y) dy. Then, if we multiply both sides of the original differential equation by I, the left-hand side will be an exact differential.

In this case, we can see that dI/I = x dx + (-y) dy, so I = e^(∫x dx + ∫(-y) dy) = e^(x^2/2 - y^2/2).

Then, the integrating factor for the given differential equation is I = e^(x^2/2 - y^2/2) = 1/xy.