 Hi and welcome to our session. Let us discuss the following question. The question says, solve the differential equation d2r by dx2 equals to x square plus sine 3x given that dy by dx is equal to 1 is equal to y when x is equal to 0. Let's now begin with the solution. In this question, we have to solve the differential equation d2y by dx2 equals to x square plus sine 3x. Now, we will integrate both sides of this equation. On integrating both sides of this equation, we get dy by dx equals to integral of x square plus sine 3x with respect to x. Now, this is equal to integral of x square with respect to x plus integral of sine 3x with respect to x. Now, integral of x square with respect to x is x2 by 3 and integral of sine 3x with respect to x is minus plus 3x divided by 3 plus c1. And dx is equal to 1 when x is equal to 0. So, by substituting 1 in case of dy by dx and 0 in case of x, we get 1 equals to 0 by 3 minus cos 0 by 3 plus c1. Now, this implies 1 is equal to minus 1 by 3 plus c1 and this implies c1 is equal to 4 by 3. So, dy by dx is equal to, equals to...