Revision Notes - Differential Equations Notes | Study Mock Test Series for JEE Main & Advanced 2022 - JEE

• The order of the differential equation is the order of the derivative of the highest order occurring in the differential equation.
• The degree of a differential equation is the degree of the highest order differential coefficient appearing in it subject to the condition that it can be expressed as a polynomial equation in derivatives.
• A solution in which the number of constants is equal to the order of the equation is called the general solution of a differential equation. 
• Particular solutions are derived from the general solution by assigning different values to the constants of general solution.
• An ordinary differential equation (ODE) of order n is an equation of the formF(x, y, y',….., y⁽ⁿ⁾ ) = 0, where y is a function of x and y' denotes the first derivative of y with respect to x.
• An ODE of order n is said to be linear if it is of the form an(x)y⁽ⁿ⁾ + a^n-1(x) y ^(n-1) + …. + a₁(x) y' + a₀ (x) y = Q(x)
• If both m₁ and m₂ are constants, the expressions (D – m₁) (D – m₂) y and (D– m₂) (D – m₁) y are equivalent i.e. the expression is independent of the order of operational factors.
• A differential equation of the form dy/ dx = f (ax+by+c) is solved by writing ax + by + c = t.
• A differential equation, M dx + N dy = 0, is homogeneous if replacement of x and y by λx and λy results in the original function multiplied by some power of λ, where the power of λ is called the degree of the original function.
• Homogeneous differential equations are solved by putting y = vx.
• Linear equation are of the form of dy/dx + Py = Q, where P and Q are functions of x alone, or constants.
• Linear equations are solved by substituting y =uv, where u and v are functions of x.
• The general method for finding the particular integral of any function is 1/ (D-α)x = e^αx∫Xe^-αxdx

Various methods of finding the particular integrals:
1. When X = e^ax in f(D) y = X, where a is a constant
Then 1/f(D) e^ax = 1/f(a) e^ax , if f(a) ≠ 0 and
1/f(D) e^ax = x^r/f^r(a) e^ax , if f(a) = 0, where f(D) = (D-a)^rf(D)
2. To find P.I. when X = cos ax or sin ax
f (D) y = X
If f (– a²) ≠ 0   then 1/f(D²) sin ax = 1/f(-a²) sin ax
If f (– a²) = 0 then (D² + a²) is at least one factor of f (D²)
3. To find the P.I.when X = xm  where m ∈ N
f (D) y = xm
y = 1/ f(D) xm
4. To find the value of 1/f(D) e^ax V where ‘a’ is a constant and V is a function of x
1/f (D) .e^ax V = e^ax.1/f (D+a). V
5. To find 1/f (D). xV where  V is a function of x
1/f (D).xV = [x- 1/f(D). f'(D)] 1/f(D) V

Some Results on Tangents and Normals:
1. The equation of the tangent at P(x, y) to the curve y= f(x) is Y – y =  dy/dx .(X-x)
2. The equation  of the  normal  at point P(x, y)  to the  curve y = f(x) isY – y =  [-1/ (dy/dx) ].(X – x )
3. The length of the tangent  = CP  =y √[1+(dx/dy)²]
4. The  length of the normal = PD = y √[1+(dy/dx)²]
5. The length of the Cartesian sub tangent  = CA = y dy/dx
6. The length of the Cartesian subnormal = AD = y dy/dx
7. The initial ordinate of the tangent = OB = y – x.dy/dx