Differential Equations HW problem. Solution and work needed, please. I have attached a picture of the section 4.1 theorem from my textbook as well. However, you can refer to the textbook if needed. It is called "A first course in differential equations the classic 5th edition" by Dennis Zill. Consider the second-order, linear, homogeneous differential equation withinitial conditions,x²y" +xy - 4y = 0y(0) = 0y'(0) = 0This initial value problem has many solutions, for instance y1 = x2 and y2 = 0. Doesthis violate the existence and uniqueness theorem for linear differential equations, fromsection 4.1? Explain. THEOREM4.1Existence of a Unique SolutionLet a,(x), a-1(x),.., a (x), a(x), and g(x) be continuous on an interval /and let a,(x) + 0 for every x in this interval. If x = xo is any point in this in-terval, then a solution y(x) of the initial-value problem (1) exists on the in-terval and is unique.EXAMPLE 1 Solution of an IVPYou should verify that the function y = 3e + e 2initial-value problem3x is a solution of theSECTION 4.1 Preliminary Theory113y" - 4y = 12x, y(0) = 4, y'(0) = 1.Now the differential equation is linear, the coefficients as well as g(x) = 12x arecontinuous, and a(x) = 1+ 0 on any interval containing x = 0. We concludefrom Theorem 4.1 that the given function is the unique solution.EXAMPLE 2Trivial Solution of an IVPThe initial-value problem3y" + 5y" - y' + 7y = 0, y(1) -0, y'(1) = 0, y"(1) = 0possesses the trivial solution y = 0. Since the third-order equation is linear withconstant coefficients, it follows that all the conditions of Theorem 4.1 are ful-filled. Hence y = 0 is the only solution on any interval containing x = 1.EXAMPLE 3 Solution of an IVPThe function y = sin 4x is a solution of the initial-value problemy" + 16y = 0, y(0) = 0, y'(0) = 1.It follows from Theorem 4.1 that on any interval containing x = 0 the solutionis unique.The requirements in Theorem 4.1 that a,(x), i = 0, 1, 2, ..., n be continu-ous and a,(x) + O for every x in / are both important. Specifically, if a,(x) = 0for some x in the interval, then the solution of a linear initial-value problem maynot be unique or even exist.

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Differential Equations HW problem. Solution and work needed, please. I have attached a picture of the section 4.1 theorem from my textbook as well. However, you can refer to the textbook if needed. It is called &#34;A first course in differential equations the classic 5th edition&#34; by Dennis Zill. Consider the second-order, linear, homogeneous differential equation withinitial conditions,x²y&#34; +xy - 4y = 0y(0) = 0y'(0) = 0This initial value problem has many solutions, for instance y1 = x2 and y2 = 0. Doesthis violate the existence and uniqueness theorem for linear differential equations, fromsection 4.1? Explain. THEOREM4.1Existence of a Unique SolutionLet a,(x), a-1(x),.., a (x), a(x), and g(x) be continuous on an interval /and let a,(x) + 0 for every x in this interval. If x = xo is any point in this in-terval, then a solution y(x) of the initial-value problem (1) exists on the in-terval and is unique.EXAMPLE 1 Solution of an IVPYou should verify that the function y = 3e + e 2initial-value problem3x is a solution of theSECTION 4.1 Preliminary Theory113y&#34; - 4y = 12x, y(0) = 4, y'(0) = 1.Now the differential equation is linear, the coefficients as well as g(x) = 12x arecontinuous, and a(x) = 1+ 0 on any interval containing x = 0. We concludefrom Theorem 4.1 that the given function is the unique solution.EXAMPLE 2Trivial Solution of an IVPThe initial-value problem3y&#34; + 5y&#34; - y' + 7y = 0, y(1) -0, y'(1) = 0, y&#34;(1) = 0possesses the trivial solution y = 0. Since the third-order equation is linear withconstant coefficients, it follows that all the conditions of Theorem 4.1 are ful-filled. Hence y = 0 is the only solution on any interval containing x = 1.EXAMPLE 3 Solution of an IVPThe function y = sin 4x is a solution of the initial-value problemy&#34; + 16y = 0, y(0) = 0, y'(0) = 1.It follows from Theorem 4.1 that on any interval containing x = 0 the solutionis unique.The requirements in Theorem 4.1 that a,(x), i = 0, 1, 2, ..., n be continu-ous and a,(x) + O for every x in / are both important. Specifically, if a,(x) = 0for some x in the interval, then the solution of a linear initial-value problem maynot be unique or even exist.

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