 Hi and welcome to the session. I'm Shashi and I'm going to help you with the following question Question says for the following differential equations find a particular solution Satisfying the given condition. This is the given differential equation and this is the given condition Let us now start with the solution Now given differential equation is x cube plus x square plus x plus 1 multiplied by dy upon dx is equal to 2x square plus x Now separating the variables in this equation. We get dy is equal to 2x square plus x upon x cube plus x square plus x plus 1 dx Now integrating both the sides of this equation. We get Integral of dy is equal to integral of 2x square plus x upon x cube plus x square plus x plus 1 dx Now we can find this integral by using this formula of integration So this integral is equal to y Now we will write y is equal to integral of 2x square plus x Now in the denominator clearly we can see these two terms have x square common So we can write these two terms as x square multiplied by x plus 1 and These two terms have one common So we can write these two terms as 1 multiplied by x plus 1 Now we will write dx as it is here Now here we will write plus c where c is the constant of integration Now this further implies y is equal to Integral of 2x square plus x upon x plus 1 multiplied by x square plus 1 Taking x plus 1 common in these two terms we get this expression Clearly we can see here. We have factorized this expression and factors of this expression are x plus 1 multiplied by x square plus 1 Now let us assume that this integral is equal to i so we can write Y is equal to i plus c let us name this expression as 1 now we know integral i is equal to integral of 2x square plus x upon x plus 1 multiplied by x square plus 1 dx Now let us consider 2x square plus x upon x plus 1 multiplied by x square plus 1 It can be written as sum of a upon x plus 1 plus bx plus c upon x square plus 1 We know to find this integral we will use integration by partial fractions here So we have considered this integral and it can be written as sum of these two fractions Now adding these two fractions by taking their LCM we get 2x square plus x is equal to a multiplied by x square plus 1 plus bx plus c multiplied by x plus 1 Now let us name this equation as equation 2 now putting x is equal to minus 1 in equation 2 We get 2 minus 1 is equal to 2a Clearly we can see if we put x is equal to minus 1 then this term will become 0 and we get this equation Now this further implies 1 is equal to 2a 2 minus 1 is 1 only Now dividing both the sides by 2 we get 1 upon 2 is equal to a Or we can simply write a is equal to 1 upon 2 Now if we put x is equal to 0 in equation 2 then we get 0 is equal to a plus c Now we will substitute this value of a in this equation and we get 0 is equal to 1 upon 2 plus c Now subtracting 1 upon 2 from both the sides of this equation We get minus 1 upon 2 is equal to c or we can simply write c is equal to minus 1 upon 2 Now if we put x is equal to 1 in equation 2 then we get 3 is equal to 2a plus 2b plus 2c Now we know value of a is 1 upon 2 and value of c is minus 1 upon 2 So we will substitute corresponding values of a and c in this equation and we get 3 is equal to 2 multiplied by 1 upon 2 plus 2v Plus 2 multiplied by minus 1 upon 2 Now simplifying this equation further we get 3 is equal to 1 plus 2b minus 1 Here 2 and 2 will get cancelled here. Also this 2 will get cancelled by this 2 Now plus 1 minus 1 is equal to 0 So Here we will cancel 1 by minus 1 and we get 3 is equal to 2v Now dividing both the sides by 2 we get 3 upon 2 is equal to b Or we can simply write b is equal to 3 upon 2 Now we will substitute corresponding values of a, b and c in this equation and we get 2x square plus x upon x plus 1 multiplied by x square plus 1 is equal to 1 upon 2 multiplied by x plus 1 plus 3 upon 2 multiplied by x plus minus 1 upon 2 upon x square plus 1 Now it can be further written as 1 upon 2 multiplied by x plus 1 plus 1 upon 2 multiplied by 3x minus 1 upon x square plus 1 Now we know integral i is equal to integral of 2x square plus x dx upon x plus 1 multiplied by x square plus 1 Now using this expression we can write this integrand as sum of these two fractions so we can write i is equal to integral of 1 upon 2 multiplied by x plus 1 plus 1 upon 2 multiplied by 3x minus 1 upon x square plus 1 dx Now this integral can be further written as integral of 1 upon x plus 1 dx multiplied by 1 upon 2 plus 1 upon 2 multiplied by integral of 3x minus 1 dx upon x square plus 1 Now first of all we will find out this integral We will use substitution method to find this integral So we can write put x plus 1 is equal to t now differentiating both the sides with respect to x we get dx is equal to dt now integral of 1 upon x plus 1 dx is equal to integral of 1 upon t dt now This integral is further equal to log t plus c1 We know formula of integration that integral of 1 upon x dx is equal to log x Plus c so here we have used this formula of integration Now substituting the value of t here we get log of x plus 1 plus c1 So we can write this integral is equal to first of all We will write this 1 upon 2 as it is now here. We will write log of x plus 1 Now we will find out this integral Now this integral can be written as Integral of 3x dx upon x square plus 1 minus integral of dx upon x square plus 1 Now this integral is equal to tan inverse x So here we will write minus sign as it is and this integral is equal to tan inverse x Now we will evaluate this integral Now multiplying and dividing by 2 we get this integral is equal to 3 upon 2 multiplied by integral of 2x upon x square plus 1 dx Now we can find this integral by using substitution method So here we can write put x square plus 1 is equal to z Now differentiating both the sides with respect to x we get 2x dx is equal to dz Now this further implies dx is equal to dz upon 2x Now substituting z for x square plus 1 and dz upon 2x for dx in this integral we get 3 upon 2 multiplied by integral of dz upon z minus tan inverse x Now this integral is equal to log of z So here we will write 3 upon 2 as it is and We get 3 upon 2 multiplied by log z minus tan inverse x Now substituting the value of z here we get 3 upon 2 multiplied by log of x square plus 1 minus tan inverse x Plus c2 where c2 is the constant of integration Now we will write this plus 1 upon 2 as it is and we know this integral is equal to 3 upon 2 multiplied by log of x square plus 1 minus tan inverse x Now simplifying this expression further we get 1 upon 2 multiplied by log of x plus 1 plus 3 upon 4 multiplied by log of x square plus 1 minus 1 upon 2 tan inverse x Now taking 1 upon 4 common in these two terms we get 1 upon 4 multiplied by 2 log of x plus 1 plus 3 log of x square plus 1 minus 1 upon 2 tan inverse x Now using this law of logarithms We can write 2 log x plus 1 is equal to log of x plus 1 whole square So here we will write 1 upon 4 as it is and We can write this term as log of x plus 1 whole square plus Now again Applying this law of logarithms in this term we can write this term as log of x square plus 1 whole cube minus 1 upon 2 tan inverse x Now using this law of logarithms We can write these two terms as log of x plus 1 whole square multiplied by x square plus 1 whole cube minus 1 upon 2 tan inverse x So this is the required value of i Now we will substitute value of i in equation 1 Now equation 1 becomes y is equal to 1 upon 4 multiplied by log of x plus 1 whole square multiplied by x square plus 1 whole cube minus 1 upon 2 tan inverse x plus c Now the given condition in the question is y is equal to 1 when x is equal to 0 So we will substitute y is equal to 1 and x is equal to 0 in this equation To find the value of c Let us name this equation as 3 Now we can write Substituting y is equal to 1 and x is equal to 0 in equation 3 We get 1 is equal to 1 upon 4 multiplied by log of 1 minus 1 upon 2 tan inverse 0 plus c Now we know log 1 is equal to 0 So we get 1 is equal to 0 minus tan inverse 0 is also 0 So 1 upon 2 multiplied by 0 is equal to 0 only Now we get 0 minus 0 plus c Now this further implies 1 is equal to c Or we can simply write c is equal to 1 Now we will substitute this value of c in equation 3 Now equation 3 becomes y is equal to 1 upon 4 multiplied by log of x plus 1 whole square multiplied by x square plus 1 whole cube minus 1 upon 2 tan inverse x plus 1 Or we can simply write y is equal to 1 upon 4 multiplied by log of x plus 1 whole square multiplied by x square plus 1 whole cube minus 1 upon 2 tan inverse x plus 1 So this is our required particular solution This completes the session. Hope you understood the solution. Take care and have a nice day