CSIR NET MATHEMATICS | Ordinary Differential Equation | Wronskian of ODE Q.92(A) Q.96(B) Q.101(C)

Hello friends, in this video Inside, we have an ordinary differential On the equation problem Describe CS Ain June 2019 example 101 Brochure C Marx Point Center y'' + a x y = 0 A x is a continuous function with the period. Let φ₁(x), φ₂(x) be the basis for the solution satisfying the initial condition phi 1 0 equals 1, phi 1 dash 0 equals 0, phi 2 0 equals 0 and phi 2 dash 0 equals Let w(φ₁, φ₂) denote the brown skin of φ₁x. and phi 2 x. Then we have four options. First option, brown skin of φ₁ φ₂ equal to 1, brown skin of φ₁ 2 equals e to the power of x Option 3, Phi 1 T plus phi 2 minus t equals 2 if the given differential equation has non-trivial periodic solution with period T and option number 4, phi 1 t plus phi 2 minus t equals 1 if the given differential equation has non-trivial solution with period t This is a problem of Vrānsken. So first, we have the second-order differential equation Vrāṃsaken form of We will recall, standard form a second-order differential equation is y double dash pxy plus qxy equal to 0 Here, the coefficient of px y dash is Thus, the Wronskian of the solutions 5152 equals c times e to the power of minus the integral of p(x) dx. And another form is the Wronskian of two functions, 5152 determinant 5152, 51 dash, 52 dash In our example, px equals 0. Because the coefficient of px y dash is And the term with y dash is not even present in our equation. So px is equal to 0 Then Vronskinn 5152 will be C E power minus PX DX Here PX 0, so C E power integration 0 DX So C E power 0 equals Then it will become constant C, Voronskin. And Anna from Vronsky's Determinant 51-52, 51-52 So 51X, 52-X, 52X, 51-X We are given the initial conditions. 4 initial conditions 510 equals 1, 51-0 equals 0, 520 equals 0, 52-0 equals 1 Using this initial condition, we will find the value of c. so substitute the initial condition ground skin of 510, 520 equal to c we will substitute into the second equation so 510, 520, 520, 510 The value of 510 is 1 The value of 520 is 1 The value of 520 is 0. And the value of 510 is 0 so 1 into 1 minus 0 into 0 So the answer will be 1 If we compare both ground skins Then the value of C will become 1. Then the answer will be 1 And the answer inside our Option 1 is 1. And the answer for option 2 is E to the power X. So option 1 is correct. Option 2 will become incorrect Now initial condition Option number 3 and Option number 4 They are talking about the periodic solution T. so a periodic solution with period T So period T is T The value will repeat after the period. So 510 plus T equals 1. 51 - 0 plus t equals 0 52 + t = 0 52 - 0 plus t equals 1 therefore 51 equals 1 51 - t equals 0 52 - t equals 0 52 - t equals 1 We need to find the sum here: 51 - t plus 52 - t so 51-t plus 52-t Both have a value of 11. So 1 plus 1 equals 2. So option number 3 51st plus 52nd equals 2 It will be correct. And option number 4 5 1 t plus 5 2 dash t equal to 1 It was wrong. So option 3 is correct and option 4 is incorrect. In this example, there are two correct answers. 1 and 3 Thank you for watching for more videos Like, share, subscribe, and press the bell icon for the next channel