DIFFERENTIAL EQUATIONS

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DIFFERENTIAL EQUATIONS : 

DIFFERENTIAL EQUATIONS Differential equations are two types. Ordinery differential equations:Equations which involves one independent varial Partial differential equations:Equations which involves morethan two independent variables

FORMATION OF DEFERENTIAL EQUATION : 

FORMATION OF DEFERENTIAL EQUATION Observe the number of orbitrary constants in the given equation (n constants) If it consists one oc differentiate the equatiion once w.r.t. x If it consists n oc differentiate it n times From these n differentialequations and given equation eliminate the n oc The obtained eqaution which is free from oc is the required de of the given equation

I ORDER AND I DEGREE DIFFERENTIAL EQUATIONS : 

I ORDER AND I DEGREE DIFFERENTIAL EQUATIONS The geneneral form of I order Idegree DE is dy/dx = f(x,y) or dy/dx = f(x,y)/g(x,y) or Mdx + Ndy = 0 Types of differential equations: Variable separable type Homogeneous equations and equations reducible to Homogeneous equations Exact equations and equations reducible to exact by use of integrating factors Linear equations and Bernoullis equations To solve above different DEs there are different methods

VARIABLE SEPARABLE : 

VARIABLE SEPARABLE The equation in which the variables are separable is called variable separable eqn To solve it separate the variables and its differential coefficients and integrate them Ex:- dy/dx = √1+y2/1+x2 = 0

HOMOGENEOUS EQUATIONS : 

HOMOGENEOUS EQUATIONS An equation in which the degree of the each term is equal Ex:- (x2 + y2)dx = 2xydy Method to solve it Substitute y = vx and dy/dx = v + x dv/dx After substitution the given equation converts into variable separable form in which variables are x,v Separate the variables and integrate it Replace ‘v’ with y/x to get the solution

EXACT DIFFERENTIAL EQUATIONS : 

EXACT DIFFERENTIAL EQUATIONS If the given equation in the form Mdx + ndy = 0 and if dM/dy = dN/dx that eqn is called exact equation Solution for exact equation y = ∫y constant Mdx + ∫(terms independent of x in N)dy = c Ex :- (2x – y + 1) dx + (2y – x – 1) dy = 0 M = (2x – y + 1) N = (2y – x – 1) dM/dy = -1 dN/dx = -1 dM/dy = dN/dx - Exact Sol:- ∫(2x – y +1) dx+ ∫ (2y – 1) dy = c ═> x2 – yx + x + y2 – y = c

EQUATIONS REDUCIBLE TO EXACT : 

EQUATIONS REDUCIBLE TO EXACT We can reduce the non-exact equation into exact by using integrating facotrs(If) There are different integrating factors. If M(x,y)dx + N(x,y)dy = 0 is a homogeneous differetial equation and Mdx + Ndy ≠ 0 then 1/Mx + Ny is an integrating factor If M(x,y)dx + N(x,y)dy = 0 is of the form y f(xy) dx + x g(xy)dy = 0 Mx – Ny ≠ 0 then 1/Mx - Ny is an integrating factor If dm/dy- dN/dx / N = f(x) then integrating factor is e∫ f(x)dx If dN/dx- dM/dy / M= f(y) then integrating factor is e∫ f(y)dy

LINEAR DIFFERENTIAL EQUATIONS : 

LINEAR DIFFERENTIAL EQUATIONS An euation of the form dy/dx + P(x) y = Q(x) is called a linear differential Solution :- Write integrating factor e∫p(x) dx Solution is given by y e∫p(x) dx = ∫Q(x) e∫p(x) dx dx + c Note:- Sometimes it is convenient to put the DE in the form dx/dy P(y)x = Q(y). In this case the solution is xe∫p(y) dxy= ∫Q(y) e∫p(y)dy dy +c

Bernoulli’s Differential equation : 

Bernoulli’s Differential equation An equation of the form dy/dx + P(x)y = Q(x)yn is called Bernoullis equation Working rule for solution:- Divide the given equation with yn Substitute y1-n = t and (1-n)y-ndy/dx = du/dx By the above substitution and simplification the given equation converts into linear differential equation in t and this can be solved by known method Finally replace t with y1-n

APPLICATION OF I ORDER AND I DEGREE DE : 

APPLICATION OF I ORDER AND I DEGREE DE Orthogonal trajectories Newton’s law of cooling Law of natural growth and decay

ORTHOGONAL TRAJECORIES : 

ORTHOGONAL TRAJECORIES Finding OTs in cartesian coordinates Form a differential equation for the given equation by eliminatin orbitrary constantas Replace dy/dx with –dx/dy Solve the obtained differential equation by using any one of the known method to get the OT of the given equation Finding OT in polar coordinates Form a DE for the given equation of the form f(r,θ,c) = 0 Replace dr/dθ with – r2 dθ/dr Solve the obtained differential equation by using any one of the known method to get the OT of the given equation