It's nice and too basic In this question jab hame pta chal gaya ki both roots are negative then obviously it's possible only when a and b both greater than zero
complex roots come in conjugates a±ib, the auxiliary eq. is (e^(ax))(Acosbx + Bsinbx) for the solution to go to 0 as x goes to infinity, a must be negative, sum of two roots is 2a and thus negative, so the coefficient of x is positive, and the product of complex conjugates is also a positive number (the square of the modulus).
Here,if we find the roots of auxiliary equation by "Sridhar Acharya" formula, and then get the CF of given equation, there occurs problem to find a and b.Why sure?
complex roots come in conjugates a±ib, the auxiliary eq. is (e^(ax))(Acosbx + Bsinbx) for the solution to go to 0 as x goes to infinity, a must be negative, sum of two roots is 2a and thus negative, so the coefficient of x is positive, and the product of complex conjugates is also a positive number (the square of the modulus).
Yes u r right but that's the logic of this question. Given condition is y tends to zero when x tends to infinity. So if we take imaginary roots then it doesn't satisfy.
complex roots come in conjugates a±ib, the auxiliary eq. is (e^(ax))(Acosbx + Bsinbx) for the solution to go to 0 as x goes to infinity, a must be negative, sum of two roots is 2a and thus negative, so the coefficient of x is positive, and the product of complex conjugates is also a positive number (the square of the modulus).
@@RahulMapariBasicMaths will satisfy for complex roots too...let c+id and c-id be 2 roots..then it's product is positive..so b positive.As y tend to 0 as x tend to infinity,c is negative( since y(x)=e^cx(A cos dx+B sin dx) ).sum of roots is 2c= -a, a is positive
@@RahulMapariBasicMaths sir pura to dekhe apka har video ka explanation zabardast rehta hai.but isme ye samajh nhi paa rahe hai ki roots direct real kese assume kr skte hai imaginary q nhi
Bcoz they have given every solution tends to zero as x tends to infinity so we have consider such type of solution and for that necessarily we have take real roots
complex roots come in conjugates a±ib, the auxiliary eq. is (e^(ax))(Acosbx + Bsinbx) for the solution to go to 0 as x goes to infinity, a must be negative, sum of two roots is 2a and thus negative, so the coefficient of x is positive, and the product of complex conjugates is also a positive number (the square of the modulus).
Good question ,good explanation
You are really a best teacher
Nice explanation sir thank you
Amazing explanation sir
It's nice and too basic
In this question jab hame pta chal gaya ki both roots are negative then obviously it's possible only when a and b both greater than zero
Excellent sir 🙏😍
Very fruitful video.... Want many more like this..... Solve questions which are a bit complicated.
Excellent explaination sir 🙏
Great sum sir... completely logical
Nice explaination
Waiting for ur classes thanks sir
very nice sir thanks for help us, please upload more vedio
मस्त लेक्चर सर...ty sm
Genius sir you are really deserve and great teacher
Thanks
Sir Thank you..... nic logic sir
Thank you very much sir
Appreciable work
Thanks
Mast question sir
Tysm sir i support u
Very good sir
Nice question
Tq sir
Sir aase hi. problem or btayi plz
Nice..
Thank you for video which is really helpful
Sir, Here you directly assumed that the roots are reals...
They may be complex...
complex roots come in conjugates a±ib, the auxiliary eq. is (e^(ax))(Acosbx + Bsinbx) for the solution to go to 0 as x goes to infinity, a must be negative, sum of two roots is 2a and thus negative, so the coefficient of x is positive, and the product of complex conjugates is also a positive number (the square of the modulus).
nice class
Thank you very much sir ❤❤❤❤
All ur posted video's are really helpful. Thank you sir.
nice
Thankyou sir ji
Here,if we find the roots of auxiliary equation by "Sridhar Acharya" formula, and then get the CF of given equation, there occurs problem to find a and b.Why sure?
Thanks sir
Thanq sir
Thank you sir
Sir test series chalu kijiye for csir net with vedio solution please
Thanks sir..
Sir ,in your concept what if roots your complex
Sir wt will happen if one of the soln of auxillary eqn is 0 and other is negative
Sir net ka 10 years ka questions papers post kijiye
If roots of auxiliary equation r repeated or imaginary ho to CF to alag ho jayega.... To us case me bhi option 1 e answer Hoga sir?
I wanted to ask the same question
complex roots come in conjugates a±ib, the auxiliary eq. is (e^(ax))(Acosbx + Bsinbx) for the solution to go to 0 as x goes to infinity, a must be negative, sum of two roots is 2a and thus negative, so the coefficient of x is positive, and the product of complex conjugates is also a positive number (the square of the modulus).
Sir, ek question tha is me...
Sir ye set exam kya hota h iske form kaise fill kiye jate h kb nikl te h plzz tell me
State eligibility Test.
Rahul sir do have any class also, if yes then Pls tell me sir
Is obc non crimileyar certificate for 1 year
Plz Sir Dec 2015/Booklet code-A/q.no.- 94 solve kar dijiye na..
Send me question
Ok Sir
Sir how we can write the complementary solution in exponential form,may be roots r imaginary
Yes u r right but that's the logic of this question. Given condition is y tends to zero when x tends to infinity. So if we take imaginary roots then it doesn't satisfy.
Jii sir, Thank you sir
complex roots come in conjugates a±ib, the auxiliary eq. is (e^(ax))(Acosbx + Bsinbx) for the solution to go to 0 as x goes to infinity, a must be negative, sum of two roots is 2a and thus negative, so the coefficient of x is positive, and the product of complex conjugates is also a positive number (the square of the modulus).
@@RahulMapariBasicMaths will satisfy for complex roots too...let c+id and c-id be 2 roots..then it's product is positive..so b positive.As y tend to 0 as x tend to infinity,c is negative( since y(x)=e^cx(A cos dx+B sin dx) ).sum of roots is 2c= -a, a is positive
Sir how u have taken roots are real.
They may be imaginary too
Plz see full video and read question carefully
@@RahulMapariBasicMaths sir pura to dekhe apka har video ka explanation zabardast rehta hai.but isme ye samajh nhi paa rahe hai ki roots direct real kese assume kr skte hai imaginary q nhi
Bcoz they have given every solution tends to zero as x tends to infinity so we have consider such type of solution and for that necessarily we have take real roots
complex roots come in conjugates a±ib, the auxiliary eq. is (e^(ax))(Acosbx + Bsinbx) for the solution to go to 0 as x goes to infinity, a must be negative, sum of two roots is 2a and thus negative, so the coefficient of x is positive, and the product of complex conjugates is also a positive number (the square of the modulus).
How to find the real valued solutions of the D.E y'''-iy"+y'-iy=0
Find roots. Write complementary solution then decidevreal
What is the difference between the{ sin(x):x€R} and {sin(n):n€N} how to find limit points can u show us by graphically
x is any real number where as n is natural number...
first set is uncountable whearas the second set is countable, they both are in bijection with Real numbers and Natural numbers respectively.
Thank you sir
Thank you sir
thank you sir