Revision: Mathematics >> Differential Equations CUET (UG) Differential Equations

An equation involving independent variable(s), dependent variable(s), derivatives of the dependent variable (s) with respect to the independent variable(s), and a constant is called a differential equation.

The order of the highest differential coefficient (or the highest order derivative appearing in a differential equation) is the order of the differential equation.

The highest exponent of the highest derivative is called the degree of a differential equation, provided exponents of each derivative and an unknown variable appearing in the differential equation are non-negative integers.

The equation \[\frac{dy}{dx}=f(x,y)\] is to be in variable separable form if it can be expressed as \[h(x)dx=g(y)dy\].

The solution to this equation is obtained by integrating h(x) and g(y) with respect to x and y, respectively.

A differential equation of the form \[\frac{dy}{dx}=\frac{f_{1}(x,y)}{f_{2}(x,y)},\] where f₁(x, y) ) and f₂(x, y)  are homogeneous functions of x and y of the same degree, is called a homogeneous differential equation.

A linear differential equation of first order and first degree is
\[\frac{\mathrm{d}y}{\mathrm{d}x}+\mathrm{P}y=\mathrm{Q}\], where P and Q are the functions of x or constants. Its general solution is  \[y.\left(\mathrm{I.F.}\right)=\int\mathrm{Q.}\left(\mathrm{I.F.}\right)\mathrm{d}x+\mathrm{c}\] and the function \[\mathrm{e}^{\int\mathrm{Pdx}}\] is called the integrating factor (I.F.) of the given equation.

Since ‘a’ lies between 0 and 2a,
we have

`int_0^(2a)f(x)dx=int_0^af(x)dx+int_a^(2a)f(x)dx,  .......(byint_a^bf(x)dx=int_a^cf(x)dx+int_c^bf(x)dx)`

`=I_1+I_2`     ........................(say)

`I_2 = int_a^(2a)f(x)dx`

Put x = 2a − t

Therefore, dx = −dt

When x = a, 2a − t = a

t = a

When x = 2a, 2a − t = 2a

t = 0

`I_2 = int_0^(2a) f(x) dx = int_a^0 f(2a - t) (-dt)`

`= -int_a^0 f(2a - t)dt = int_0^a f(2a - t)dt      ...................... (By int_a^b f(x)dx = -int_b^a f(x)dx)`

`=int_0^a f(2a - x)dx    ..............(By int_a^b f(X)dx = int_a^b f(t)dt)`

`int_0^(2a) f(x)dx = int_0^a f(x)dx + int_0^a f(2a - x)dx`

`= int_0^a [f(x) + f(2a - x)]dx`

To show that:

`int_0^pi sin x  dx = 2 int_0^(pi/2) sin x  dx`

We use the proven property by setting f(x) = sin x and 2a = π, which means a = `pi/2`.

The property tells us that:

`int_0^pi sin x  dx = int_0^(pi/2) sin  x  dx + int_0^(pi/2) sin (pi - x)  dx`

Knowing the trigonometric identity sin (π - x) = sin x, the equation simplifies to:

`int_0^pi sin x  dx = 2 int_0^(pi/2) sin x  dx`

This directly applies the property to the integral of sin x over [0, π] to show it equals twice the integral of sin x over `[0, pi/2]`, demonstrating the utility of this property in simplifying integrals with symmetric functions over specific intervals​.

(i) Express the homogeneous differential equation in the form
dy/dx = f(x, y) / g(x, y)

(ii) Put y = vx and
dy/dx = v + x dv/dx

Substitute in the equation and cancel out x from the R.H.S.
The equation reduces to the form
v + x dv/dx = F(v)

(iii) Take v on R.H.S. and separate the variables v and x

(iv) Integrate both sides to obtain the solution in terms of v and x

(v) To obtain the required solution in terms of x and y, substitute v = y/x

(i) Write the equation in the form dy/dx + Py = Q

(ii) Identify P and Q

(iii) Find I.F. = \[\mathrm{e}^{\int\mathrm{Pdx}}\]

(iv) Multiply the whole equation by I.F.

(v) Integrate and get a solution.