CSIR NET MATHEMATICS | Ordinary Differential Equation | Homogeneous Equations June 2019

Hello friends, in this video we are going to discuss the problem of ordinary differential equation from CSINE June 2019. Example number 48, booklet c, marks 3. The positive values of lambda for which the equation y double dash plus lambda square y equal to zero has non-trivial solution satisfying y of zero is equal to y pi and y dash of zero is equal to y dash of pi are the first option lambda is equal to 2n plus 1 divided by 2 for the natural number n second option lambda equal to 2n third option lambda equal to n and fourth option 2n minus 1 now the given differential equation is a homogeneous linear differential equation therefore auxiliary equation for the given ode is m square plus lambda square equal to 0 so factorization m minus i lambda plus m plus i lambda equal to 0 then we have two values of m m1 equal to i lambda plus m2 equal to minus i lambda these are the complex conjugate so corresponding solution is of the form c1 cos lambda x plus c2 sine lambda x now we again condition y of 0 is equal to y pi and y dash of 0 is equal to y dash of pi our solution is y equal to c1 cos lambda x plus c2 sine lambda x if we substitute x is equal to 0 then we have c1 cos 0 which has value 1 plus c2 sine 0 which has value 0 on the right hand side x is equal to pi so c1 cos lambda pi plus c2 sine lambda pi then finally we have equation c1 equal to c1 cos lambda pi plus c2 sine lambda pi first equation for the second condition we required derivative of y so if we differentiate y with respect to x then we have minus lambda c1 sine lambda x plus lambda c2 cos lambda x and the second condition is y dash of 0 is equal to y dash of pi so left hand side just substitute x is equal to 0 sin 0 0 plus cos 0 1 for the right hand side x is equal to pi so minus lambda c1 sin lambda pi plus lambda c2 cos lambda pi now if we eliminate lambda from the both side we have c2 minus c1 sine lambda pi plus c2 cos lambda pi now from these two equation we try to eliminate cos lambda pi or sine lambda pi to find the value of cos lambda pi or sine lambda pi so for that we multiply first equation by c1 and second equation by c2 then we have first equation c1 square is equal to c1 square cos lambda pi plus c2 c1 sine lambda pi now in the second equation c2 square equal to minus c2 c1 sine lambda pi plus c2 square cos lambda pi so plus c2 c1 sine lambda pi cancel with the minus c2 c1 sine lambda pi when we add this to equation and we have c1 square plus c2 square left hand side and right hand side c1 square cos lambda plus c2 square cos lambda now both the side we have c1 square plus c2 square if c1 c2 0 then c1 square plus c2 square 0 but we want non-trivial solution therefore this c1 square plus c2 square never be 0 so if we divide both side by c1 square plus c2 square then we have cos lambda pi equal to 1 and we know lambda pi equal to 2 n pi then cos lambda pi will be 1 and if we remove pi both side then we lambda equal to 2n and which is the option number 2 therefore option number 2 is a correct answer thanks for watching for more videos like share subscribe press the bell icon of next gen maths