CSIR NET MATHEMATICS| Ordinary Differential Equations | Properties General Solution & Solution

Hello friends, in this video we are going to discuss the problem of ordinary differential equation from the CSIR NET December 2018. Example number 99, booklet C which is a problem of marks 4.75. Three solutions of certain second order non-homogeneous linear differential equation are y1x is equal to 1 plus x into e power x square y2x is equal to 1 plus x into e power x square minus 1. y3x is equal to 1 plus e power x square. Which of the following is or are general solution of differential equation? The first option c1 plus 1 y1 plus c2 minus c1 y2 minus c2 y3 where c1 c2 are arbitrary constants. Now the second option, c1 y1 minus y2 plus c2 y2 minus y3 where c1 c2 are the arbitrary constant. Third option, c1 y1 minus y2 plus c2 into y2 minus y3 plus c3 into y3 minus y1 where c1 c2 c3 are arbitrary constant. And the fourth option, c1 into y1 minus y3 plus c2 into y3 minus y2 plus y1 where c1 c2 are arbitrary constant. Now, we begin with the general form of second order non-homogeneous linear differential equation which is d square y by dx square plus dy by dx plus qy equal to fx. If fx is equal to 0, then it will convert into the homogeneous differential equation but we have a non-homogeneous differential equation. Therefore, f of x should not be 0. Now, the solution As we know the second order linear differential equation has infinitely many solution but two solution should be independent and the general solution contain only two arbitrary constant. Now from this second property of the solution of second order non-homogeneous differential equation the third option which contain three arbitrary constant so that should be wrong. So, first we discard the third option which contains three arbitrary constants. Now, suppose u1 and u2 are any two linearly independent solutions of second order non-homogeneous linear differential equation That means u1 and u2 must satisfy the second order differential equation. That means u1 and u2 must satisfy Therefore, you can write down d square u1 upon dx square plus du1 upon dx plus q u1 equal to fx and similarly u2 also satisfy the differential equation d square u2 divided by dx square plus p d u2 divided by dx plus q u2 equal to fx. General solution of the form k1 u1 plus k2 u2 also satisfy the second order non-homogeneous linear differential equation where k1 and k2 are the arbitrary constant. Therefore we have d square k1 u1 plus k2 u2 divided by dx squared. d of k1 u1 plus k2 u2 divided by dx q into k1 u1 plus k2 u2 equal to fx. Now separating terms of k1 and k2 we have the differential equation k1 into d square u1 upon dx square plus p du1 dx plus q u1 plus k2 into d square u2 upon dx square plus p into du2 by dx plus q u2 equal to fx. But this is already we prove that it has value fx and this bracket has value fx. Therefore, we have k1 into fx plus k2 into fx is equal to fx. As we discussed, it is a non-homogeneous differential equation. So, fx should be non-zero. If it is non-zero, so divide both this side by fx. Then we have k1 plus k2 equal to 1. So, the general solution k1u1 plus k2u2 will be the general solution if k1 plus k2 equal to 1. Now, we are going to check all the option one by one. If the option satisfy this condition k1 plus k2 equal to 1, then we can say that it is a solution of the given differential equation which has y1, y2, yn are the solution. Option 1 The arbitrary constant As a coefficient of solution y1, y2, y3 are c1 plus 1, c2 minus c1 and minus c2. If we take the sum of these three coefficients c1 plus 1 plus c2 minus c1 minus c2 is equal to 1 which satisfy this condition therefore 1 is a correct answer. Now for the second option. c1 into y1 minus y2 plus c2 into y2 minus y3. Now, we rewrite the solution c1 y1 plus c2 minus c1 y2 minus c2 y3. Now, take the sum of coefficient of y1 y2 y3. We have c1 plus c2 minus c1 minus c2. Therefore, value is 0 and this does not satisfy the condition k1 plus k2 equal to 1. Hence, the second option is wrong. Now option number 3, we have given option 3, c1 y1 minus y3 plus c2 y3 minus y2 plus y1. We write in the form c1 plus 1 y1 minus c2 y2 plus c2 minus c1 y3. So, the coefficient of y1 is equal to c1 plus 1 coefficient of y2 is minus c2 and the coefficient of y3 c2 minus c1. If take the sum of these three coefficient then we have c1 plus 1 minus c2 plus c2 minus c1 has value Hence it also satisfy the condition k1 plus k2 equal to 1. Therefore the fourth option is This is also correct. So, this example has two correct answers first and fourth. Thanks for watching. Like, share, subscribe and press the bell icon of next gen.