CSIR NET MATHEMATICS |Ordinary Differential Equation|Complete Concept,Theory & Solution June 2018

In this video, we are going to discuss the problem of Ordinary Differential Equation from the CSIRnet, June 2018. Example number 92, from booklet A, the problem is of mark 4.75. Consider the system of differential equation, dx by dt is equal to 2x minus 7y, dy by dt is equal to 3x minus 8y. Then the critical point zero zero of the system is an First option. Asymptotically stable node, unstable node, asymptotically stable spiral or unstable spiral. Before solving this example, we are going to understand the theory of the differential equation, a system of differential equation. The system of linear differential equation of the form dx dx by dt is equal to ax plus by, dy by dt equal to cx plus dy which is the given form of the differential equation. Now, matrix form dx by dt dy by dt equal to a, b, c, d, x, y. The whole system can be written in the matrix form x dash is equal to a x where x dash is dx by dt dy by dt vector. a is equal to matrix a, b, c, d and x is equal to x, y. Now, this matrix system has critical point or equilibrium point. So, first we are going to define the critical point of the x dash is equal to a x system. The critical point of the equilibrium point x zero y zero of the system of differential equation is a point where x dash is equal to 0. So x dash equal to 0 that means x dash is equal to 0 and y dash is equal to 0. For the given example x dash is equal to 0 that means 2x minus 7y equal to 0 and y dash is equal to 0, therefore 3x minus 8y is equal to 0 Now, by solving these two equations, we have value of x and y, x is equal to 0, y equal to 0. That's why 0, 0 is the critical point of the system, which is already given in the example. Now, classification of critical point. The classification of critical point x0, y0 depends on the eigenvalues of which are the roots of Determine A minus lambda equal to zero that means lambda square minus trace of A into lambda plus determinant A equal to zero. So this is a characteristic equation of matrix A and the classification of critical point is depend on the lambda 1 lambda 2. It is a quadratic equation so we have only two eigenvalues and as we know for the quadratic equation varies we have three Three possibilities for these solutions. Real and distinct, second one is real and equal and third one is complex conjugate. So, we are working on that part. So, we have case 1, distinct real eigenvalues. First case, if both have the same sign, then the critical point is called the node point. Both the eigenvalues, lambda 1, lambda 2 are negative, then the critical point is asymptotically stable node. If both are the positive values, then the critical point is unstable. Now the second part suppose both have the opposite sign that means one is positive another is negative then the critical point is called the saddle point and which is always unstable. Now the case 2 where we have equal real eigenvalues or repeated eigenvalues now for that we require the linearly independent eigenvectors. Suppose the eigenvalue has two linearly independent eigenvectors. Then the critical point is called the proper node or called the star point. If the eigenvalue is negative, then it is a asymptotically stable proper node. And if eigenvalue is positive, then the critical point is unstable proper node. Now, in the second case, where the eigenvalue has only one linearly independent eigenvector, then the critical point is called the improper node. If the eigenvalue is negative then critical point is asymptotically stable improper node and if it is positive then it is unstable improper. Case 3 suppose the eigenvalues are complex conjugate then the real part of eigenvalue are zero that means it is purely imaginary then the critical point is called the center and the center is a stable but not asymptotically stable and suppose the real part of eigenvalues are non-zero then the critical point is called the spiral point. If the real part of the eigenvalue is negative then the critical point is called asymptotically stable spiral point. If the real part is positive then the critical point is unstable spiral point. Now for the given example the matrix Example for the given system of linear equation is dx by dt equal to 2x minus 7y and dy by dt equal to 3x minus 8y. So we can rewrite in the form of matrix dx by dt dy by dt equal to matrix 2 minus 7 3 minus 8 xy. Then we have x dash is equal to ax where a is equal to 2 minus 7 3 minus 8. Now we are interested in the eigenvalues of matrix A. So characteristic of matrix det(lambda I minus A) equal to 0. So, determinant lambda minus 2, 7, minus 3, lambda plus 8 equal to 0. Then, we have the characteristic equation lambda square plus 6 lambda minus 16 plus 21 equal to 0. So, characteristic equation is lambda square plus 6 lambda plus 5 equal to 0. And, the roots are lambda plus 1, lambda plus 5 equal to 0. Therefore, lambda 1 and the lambda 2 are minus 1, minus 5. which are the distinct eigenvalues and both have the same sign negative sign so it is called the node and they are negative that therefore it is a asymptotically stable node. Hence this example as the first option is correct asymptotically stable node. Thanks for watching for more videos like, subscribe, Please subscribe, press the bell icon of NextGenMaths.