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

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

The document Revision Notes - Differential Equations Notes | Study Mock Test Series for JEE Main & Advanced 2022 - JEE is a part of the JEE Course Mock Test Series for JEE Main & Advanced 2022.
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• 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) + an-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 = eαx∫Xe-αxdx

Various methods of finding the particular integrals:
1. When X = eax in f(D) y = X, where a is a constant
Then 1/f(D) eax = 1/f(a) eax , if f(a) ≠ 0 and
1/f(D) eax = xr/fr(a) eax , 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) eax V where ‘a’ is a constant and V is a function of x
1/f (D) .eax V = eax.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

The document Revision Notes - Differential Equations Notes | Study Mock Test Series for JEE Main & Advanced 2022 - JEE is a part of the JEE Course Mock Test Series for JEE Main & Advanced 2022.
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