Assume that y is the solution of the initial-value...

Assume that y is the solution of the initial-value problem y' + y = { 3 sin x / x, x ? 0; 3, x = 0 }, y(0) = 1 . If y is written as a power series y = ? (n=0 to ??) c_n x^n , then y = 1 + [ ] x + [ ] x^2 + [ ] x^3 + [ ] x^4 + . . . . Note: You do not have to find a general expression for c_n . Just find the coefficients one by one.

Transcript

00:01 In this question we have been given that y is the solution okay so y is given to be the solution of given equation initial value problem which is y -daz plus y it is equals to 3 sine x over x when x is non -zero and it is equals to 3 when x equals to 0 with the initial condition y 0 equals to 1 okay so we have to write y in terms of the power series cn x raised to n and going from zero to infinity.

00:36 So let us see how we can do this.

00:40 So to solve what we are going to do is the following.

00:46 So we consider this whole thing as f of x so we get f of x equals to 3, sine x over x and this is 3.

00:57 This is when x not equals to and when x equals to zero clearly this is a continuous function so clearly f of x is continuous why this is continuous we can see from this limit extending to zero f of x if you evaluate you will get limit extending to zero three sine x over x is just one so three times one equals to three only so limit exists and it is equal to the value of the function which is f of zero which is given to be three so this is continuous okay also we have this relation that y at the zero equals to one okay so here what we are going to use f of zero it is f of x equals to three okay f of x equals to three we are going to make use of now the problem problem reduces to y -daz plus y equals to three only.

02:12 So this is nothing but t -y over d x plus y equals to three.

02:18 This is a linear differential equation.

02:21 Okay, so to solve this, we need to find out the integrating factor.

02:26 So from this, if i write down my integrating factor, which is a coefficient of y.

02:33 Okay so this is given by e raise to integral the coefficient of y which is one here times b x so we get e raise to x only now we'll try to find out what is the general solution okay so let's say this is our equation number one so the general solution of this equation will be given by y times integrating factor equals to integral of the right -hand side times integrating factor all integral.

03:12 Okay, so this is what we get...

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