 Hi and how are you all today? My name is Priyanka and I shall be helping you with the following question It says solve the following differential equation given that y is equal to 1 when x is equal to 2 Now here we are given x dy by dx plus y is equal to x cube as the differential equation Let us proceed by solving this differential equation Now first of all let us divide both the sides by x on doing so we have dy by dx plus y upon x equal to x cube upon x that is equal to x square now This is a linear differential equation of the form dy by dx plus p y is equal to q, right? So on comparing these two differential equation we have the value of p as 1 by x and The value of q as x square Now what we need to do next is we need to find out the integrating Factor that is equal to e raised to the power integral p dx that is e raised to the power p is 1 by x dx that is e raised to the power log x that is in turn the value of Integrating factor is coming out to be x now the required solution is y into integrating factor equal to integral q into integrating factor dx plus c that gives us x y is equal to Integral x cube dx plus c That is x into y is equal to x raised to the power 4 upon 4 plus c now We know that It's given to us basically that y is 1 when x is equal to 2 so on substituting these two values in this we have 2 into 1 equal to x raised to the power 4 that is 2 raised to the power 4 upon 4 plus c that gives us 2 is equal to 16 upon 4 plus c that further implies the value of c is equal to minus 2 So the required answer to this question is What we need to do is we will be substituting the value of c as minus 2 Inher and we have it as x y equal to x raised to the power 4 upon 4 minus 2 as the required Solution of the given differential equation, right? So hope you understood it well and enjoyed it to have a nice day ahead