Revision: Calculus >> Differential Equations Maths Commerce (English Medium) Class 12 CBSE

A differential equation that contains ordinary derivatives of one or more dependent variables with respect to a single independent variable is called an ordinary differential equation.

Example:

If an equation contains derivatives of one dependent variable with respect to one or more independent variables, then it is called a differential equation.

Example

This is a differential equation because it contains the derivative \[\frac{dy}{dx}\].

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

A solution of a differential equation in which the number of arbitrary constants equals the order of the differential equation is called the general solution of the differential equation.

Any relation between independent and dependent variables which does not involve derivatives, such that this relation and the derivatives obtained from it satisfy the given differential equation, is called a solution of the differential equation.

A solution obtained from the general solution by giving particular values to the arbitrary constants is called a particular solution.

A differential equation \[ \frac{dy}{dx} = F(x,y) \] is said to be in variable separable form if \[ F(x,y) = g(x)h(y). \]

Thus, \[ \frac{dy}{dx} = g(x)h(y) \] and, if \[ h(y) \neq 0, \]

\[ \boxed{\,\frac{1}{h(y)}\,dy = g(x)\,dx\,}. \]

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 function \[ F(x,y) \] is called homogeneous of degree \[ n \] if

\[ \boxed{\,F(\lambda x, \lambda y) = \lambda^{n} F(x,y)\,} \]

for any non-zero constant \[ \lambda. \]

A first-order and first-degree differential equation of the form

\[ \boxed{\,\frac{dy}{dx} + Py = Q\,} \]

where \[ P \] and \[ Q \] are constants or functions of \[ x \] only, is called a first-order linear differential equation.

Similarly, if \[ x \] is treated as a function of \[ y, \] it may be written as

\[ \boxed{\,\frac{dx}{dy} + P_{1}x = Q_{1}\,} \]

where \[ P_{1} \] and \[ Q_{1} \] are constants or functions of \[ y \] only.

The factor e^∫P dx on multiplying by which the left-hand side of the differential equation\[\frac{dy}{dx}+Py=Q\] becomes the differential coefficient of a function of x and y, is called the integrating factor of the differential equation.

General Solution: \[y\cdot\mathrm{I.F.}=\int Q\cdot\mathrm{I.F.}dx+c\]

A differential equation is non-linear if any one of the following holds:

1. The degree is more than one

2. Any differential coefficient has an exponent of more than one

3. Exponent of the dependent variable is more than one

4. Products containing the dependent variable and its differential coefficients are present

A first-order differential equation, along with an initial condition, is called an initial value problem.

For y = vx:

\[ \boxed{\,\frac{dy}{dx} = v + x\,\frac{dv}{dx}\,} \]

\[ \frac{dv}{g(v) - v} = \frac{dx}{x} \]

For x = vy:

\[ \boxed{\,\frac{dx}{dy} = v + y\,\frac{dv}{dy}\,} \]

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​.

• A differential equation contains derivatives.

• An ordinary differential equation contains derivatives with respect to only one independent variable.

• Differential equations describe rates of change in mathematics and science.

• Order = highest derivative order.

• Degree = power of highest derivative.

• Degree exists only for polynomial equations in derivatives.

• Always check polynomial condition before stating the degree.

• Order and degree, if defined, are always positive integers.

• A differential equation contains derivatives of an unknown function.

• Its solution is generally a function, not a single number.

• The graph of the solution function is called the solution curve or integral curve.

• A general solution contains arbitrary constants.

• A particular solution is obtained by assigning fixed values to those constants.

• To verify a solution, substitute the function and its derivatives into the equation and check whether LHS = RHS.

• A variable separable equation can be rearranged so that all y-terms are with dy and all x-terms are with dx.

• Separate variables first, then integrate.

• Use one constant of integration.

• Apply the initial condition only after getting the general solution.

• Final answers may be explicit or implicit.

• Check homogeneity first.

• Differentiate substitution carefully.

• Convert to separable form.

• Back-substitute to original variables.

1. Write the equation in the form dy/dx + Py = Q
2. Identify P and Q or P₁ and Q₁
3. Find I.F. = 

   \[ \frac{dy}{dx} + Py = Q \Rightarrow \text{I.F.} = e^{\int P\,dx} \]

   \[ \frac{dx}{dy} + P_{1}x = Q_{1} \Rightarrow \text{I.F.} = e^{\int P_{1}\,dy} \]

4. Multiply the whole equation by I.F.
5. Integrate and get a solution.

• Write the equation in the form
  \[\frac{dy}{dx}+Py=Q\]

• Find the integrating factor
  \[\mathrm{I.F.}=e^{\int Pdx}\]

• Multiply the entire equation by I.F.

• Integrate both sides w.r.t x

• Obtain
  \[y(\mathrm{I.F.})=\int Q(\mathrm{I.F.})dx+c\]

1. Radioactive Decay:  \[x=x_0e^{-kt}\]

2. Half-Life Formula: \[k=\frac{\ln2}{T}\]

3. Newton’s Law of Cooling: \[\theta=\theta_0+(\theta_1-\theta_0)e^{-kt}\]

4. Population Growth: \[P=ae^{kt}\]