d^2y/dx^2-(cotx)dy/dx-(1-cotx)y=e^xSinx #Reductionoforder L945

Показать описание

#explanationinenglish #reductionoforder #mathspulse

Hello, People!

Here is a video of a differential equation of order two, we will solve this equation using reduction of order method. Please watch with patience and have a happy learning.

My heart felt thanks to all the subscribers, audience, supporters and well-wishers❤

With Love,
Chinnaiah Kalpana🌷

Note:

* Working rules to find part of C.F. of d^2y/dx^2+Pdy/d+Qy=0 by inspection:

1. y=e^x is the part of C.F. if 1+P+Q=0.
2. y=e^-x is the part of C.F. if 1-P+Q=0.
3. y=e^ax is the part of C.F. if a^2+Pa+Q=0.
4. y=x^m is the part of C.F. if m(m-1)+Pmx+Qx^2=0.
5. y=x is the part of C.F. if P+Qx=0.
6. y=x^2 is the part of C.F. if 2+2Px+Qx^2=0.

* Working rule to solve d^2y/dx^+Pdy/dx+Qy=R

1. Reduce the given equation to standard form as d^2y/dx^2+Pdy/dx+Qy=R-----(1)

2. Either a part of C.F. is given or it can be found by inspection in some cases given above.
Let it be y=u.

3. Let y=uv be the general solution of (1) where v=f(x)

d^2y/dx^2=u.d^2v/dx^2+(du/dx)(dv/dx)
+(du/dx)(dv/dx)+vd^2v/dx^2
=vd^2u/dx^2+2du/dcdv/dx+ud^2v/dx^2

5. With these substitutionsin(1), takes the form:
d^2v/dx^2+P1 dv/d=R1 where P1=P+2/u du/dx nd R1 = R/u

6. Taking dv/dx=V, it becomes dV/dx+P1 V=R1 which is a linear in V and x and hence it can be solved for V.

7. V=dv/dx on integration gives v.

8. Then the required general solution is y=uv.

" But in our problem, we are finding v directly using the equation
d^2v/dx^2+[P+2/u du/dx]dv/dx = R/u".

For more such videos👇

Find me on Instagram 🌼

Stay tuned to 'Maths Pulse'.

Get rid of 'Maths Phobia'.

Have a happy learning!

#chinnaiahkalpana #differenialequations #engineeringmathematics #bscmaths #math #differentialequationwithnonconstantcoefficients #problemsonreductionoforder