CSIR NET MATHEMATICS | Ordinary Differential Equations | Non-Linear ODE Q.91(A) Q.95(B) Q.101(C)

Hello friends, inside this video we will be discussing ordinary differential equations. We will discuss the problem from CSNAed June 2019. Example Number One Hundred from Booklet C Marks Four Point Seventy Five Let y be any non-trial A real solution of y'' + x y = 0. 0 < x < ∞ And let y = two x be the solution of y double prime plus y equals x squared plus two. With the condition y(0) = y'(0) = 0, then. We have four options; the first option is y one s infinitely many zeros, second option y squared minus infinitely many zeros third option y to the power of one s finitely many zeros The fourth option is Y2S finitely many zeros. Inside this example, we have Solve two ordinary differential equations, one nonlinear and the other second-order linear. So we will start with nonlinear differential equations. The solution of Y squared plus X Y equals zero For it, we will use the power series method. So, first assumption: y equals a zero plus a one x Plus A times 2x cubed plus A times 3x cubed equals two sigma equals The solution to the series sum from n = 0 to infinity of n times x to the power n is x over (1 - x) squared. Find the derivative of y dash The derivative of A zero will be zero, the derivative of A one x will be A one. The derivative of e to the 2x will be 2e to the 2x and continue Here, first we are starting from one, so The equation will become n equals one to infinity. And Deryti will be E N A times X over E N minus one Y double dot X will become zero, and F double dot will become 2A. Three A times three A's derivative will become six A three X and Continuity is starting here for the first time Equation will become n equals 2 to infinity And the derivative of x^n times x^(n-1) is will go to n(n - 1)x^(n - 1) We will substitute this value into the differential equation. y double dash plus x y equals zero The value of Y double dash is sigma n equal to two two Infinity times N into N minus one times X over N minus two And x into y, the value of y is sigma n equal to two To infinity and beyond, and to infinity and one, and to infinity and minus one, and to infinity and x over infinity and minus two Multiply this x inside, we get n x over n plus one In the next step, we have the power of (n minus 2) and (n plus 1) Convert using the same variable, so n minus To convert into K, N, and N plus one convert into We will make both of them cross the same. So first, we will replace n minus 2 with k, and add means n equals 2k + 2, so here we will write n = k + 2, k + 2 = Two times k equals zero, equation starts when k equals zero Here n equals k plus 2k plus 2 minus 1, so Here, k plus one will come, a sub n minus by a sub k plus And we will replace N mines to K with X x to the power of k replaces power and mines two In the second equation, we will replace n plus one with k, so n will become k minus one. So here, k minus 1 equals zero, so k equals one. That equation starts from k equals one to infinity. An replace by a minus one and x power k equals zero Now the power is the same here, we're also trying to make the emission the same here, the emission Starts at k equals zero, and here the session starts at k equals one We will separate the term equal to zero. So here, equal to zero zero plus two two zero plus one one E equals zero so O Zero plus two so a two will come and x power zero Equal to one, so first we will have two A two And the Riemann equation will start from equal to one to infinity plus two k plus one a plus 2ax power Both are also starting with the same number from the power of review. If you add both of them, after addition I will get k plus 2k plus 1. a k plus 2 a k minus 1 x power of Equal to zero, so the right-hand side is zero. If we take the left-hand side of the pole, then both of us are Compare the coefficients. So, comparing the coefficients, 2a² equals zero. and these coefficients are k plus 2 k plus 1 a k plus 2 Plus a k minus one equals two c Then a minus 2 equals zero will be obtained, and a plus 2 equals minus a will be obtained. Minus one plus one plus two starts here at one to infinity We will find the first value equal to one, so whatever is obtained will be the result. A plus two equals A three A plus two makes A three, A one minus one so A zero, one plus one to One plus two times three, so three equals minus zero divided by two times three. Similarly, a four will become a one in the form of one. A over three fourths negative term, A five will be obtained minus A two divided by four into five. But if it's A20, then it becomes A50. Six will come in the form of three. It will come in the final as a six-zero format. A to one, A one in form, and A two, If A5 is in the form of A5 zero, then A8 will also become zero. We have a power solution: A Zero plus A One x plus 2x cubed plus 3x cubed These values and all of them If substituted, the value of the contraction After substituting, we will get this series where Two different types of terms will appear, with a zero coefficient and with a one coefficient If we separate both terms, it will be a-zero. one minus one over two times three x cubed and continue and even will come in x minus one over three into four x to the fourth and continue Why one This is the solution to the differential equation. So, Y1 has two options. First bracket Or the second bracket, so y₁ has infinitely many zeros. For y one, I have zero possibility infinitely. The one that is our first option will become correct, and the fourth one The option says that Y1 has infinitely many zeros of color. Now the second-order differential equation y double prime plus y equals x squared plus 2 It is a linear differential equation, so first we will find the solution of its homogeneous part. So, m squared plus one equals zero, so m equals plus and minus i The solution of the Mozinis part is y equals c one cos x plus c two sin x For the non-mozinic part solution, we WYP Then y will be equal to (x squared plus one) times y equals two x We will re-ride this differential equation: square plus two So, you'll get one upon D.K. form D square plus one x square plus two, so that becomes our y prime. One upon one plus d squared, x squared plus two So we will rewrite one plus d squared over minus one x squared plus binomial theorem, and we apply it One minus d squared plus d over four continuous, d squared plus two We will multiply this operator, so x squared plus Two D squared into X squared plus two D to the power four x squared plus two minus. d squared x squared plus two, so d Derivative with respect to x of the second derivative of b with respect to x. If x is multiplied by two, then the second-order derivative of x squared is two. And the derivative of two will be zero, so the value of d squared x squared plus two will be two. And for higher-order derivatives, the answer will be zero, so here I have It will be x squared plus two minus two, so plus two minus two will cancel out. It will remain just x squared, so y equals x squared. The general solution will be obtained as y equals 2y_H plus y PYH is C1 times cos x plus C2 times sin x and y p x squared c one cos x plus c two sin x Plus x squared. Here we are given the initial condition y Equal to zero, y equals zero, equal to zero, so we will use Before that, we will find Y dash F, Y dash F is equal to You are minus C one sine x plus C two cos x plus 2x Initial condition with zero equal to zero, so put zero here X is equal to zero cosine zero is value one sine zero is value If the value of zero and x squared becomes zero, only C1 will remain. Then C1 will become equal to 2C0. Second initial condition wide is zero equal to zero here we will keep zero If x is zero, then sin zero is zero, cos zero is one, so only C2 will be obtained, and 2x as well. It will become zero, so C2 equals zero. C1 will become zero. C2 equals zero. Then the value of y will be x squared. And Y2 The solution to this differential equation is y to be It will be possible to square x and square x. Only two zeros are coming, and they are equal zeros, so y2 is finite D. Mani, the correct answer is number four, and Option number two is a dream So two wrong and fourth correct answer to this example I have two correct answers: first and fourth Thanks for watching And the peacock videos like share subscribe press the bell icon of next generation