 Hello and welcome to the session. In this session we will discuss equation reducible to variable separable. Now first of all let us discuss the differential equations with variable separable. Now a first order first degree differential equation is of the form dy over dx is equal to f of xy. Now if f of xy is expressed as the product g of x, h of y where g of x is the function of x and h of y is the function of y then this differential equation relative to the equation number one is said to be a variable separable type. Therefore the equation number one will be of the form dy over dx is equal to now f of xy is expressed as the product g of x, h of y. So dy over dx will be equal to g of x into h of y. Now if h of y is not equal to 0 then separating the variables the equation that is this equation that is equation number two can be written as 1 over h of y into dy is equal to g of x into dx and now we have separated the variables. Now integrating both sides we get integral 1 over h of y into dy is equal to integral g of x into dx. Therefore this equation provides the solutions of the given differential equation in the form h of y is equal to g of x plus c where h of y and capital g of x are the n g derivatives of 1 over h of y and g of x respectively and c is the arbitrary constant. So this is how we can find the solution of the differential equations with variables separable and now let us discuss the equation reducible to variables separable. Now sometimes in the differential equation the variables are not separable. So in this case we change the variables by using a suitable substitution and then solve the equation by the method of separation of variables. Now let us take the given differential equation is of the form dy over dx is equal to f of ax plus dy plus c the whole. That is this is the differential equation in which the variables are not separable. So for these type of differential equations put ax plus dy plus c is equal to z then change the variables that that is by putting ax plus dy plus c is equal to z in this equation. This given differential equation will be so expressed that the coefficient of dx is only the function of x and the coefficient of dz will be only the function of z. So we can easily solve that equation by using the method of separation of variables which we have discussed earlier. Now let us discuss it with the help of an example here solve dy over dx is equal to 2x plus 3y plus 1 whole square. That is we have to solve this equation in which the variables are not separable. Now we will start with its solution. Now in the first step putting 2x plus 3y plus 1 is equal to z in this equation. That is equation number one we get dy over dx is equal to z square. Now let this be equation number two. Now differentiating the equation number two with respect to x we get 2 plus 3 into dy over dx is equal to dz over dx. Which implies dy over dx is equal to 1 by 3 into dz by dx minus 2 the whole. Now let this be equation number three. So putting dy over dx is equal to 1 by 3 into dz over dx minus 2 the whole in equation number three. We get 1 by 3 into dz over dx minus 2 the whole is equal to z square. Which implies dz over dx is equal to 3z square plus 2 which further implies dz over 3z square plus 2 is equal to dx. So here we can see that the given differential equation is reduced to the differential equation of the time in which the variables are separable. Further this implies dz over taking 3 common in the denominator it will be 3 into z square plus 2 by 3 the whole is equal to dx. Which further implies dz over z square plus 2 by 3 can be written as the root 2 by 3 whole square is equal to 3 into dx. Now integrating both sides we get integral dz over z square plus root 2 by 3 whole square is equal to 3 into integral 1 into dx. Now we know this result that is the integral dx over x square plus a square is equal to 1 over a tan inverse x over a plus c. Now here a is equal to root 2 by 3. So by using this result this implies 1 over a that is 1 over root 2 by 3 tan inverse z over root 2 by 3. Is equal to 3 into x plus c. Now here c is an arbitrary constant and we will add this arbitrary constant on one side. Otherwise the solution obtained will not be general. Now this implies root 3 by 2 tan inverse z over root 2 by 3 is equal to 3x plus c which further implies. Now multiplying throughout by root 2 by 3 this will be tan inverse root 3 by 2 into z is equal to root 2 by 3 into 3x plus root 2 by 3 into c. This further implies tan inverse root 3 by 2 into z is equal to root 6 into x plus a where a is equal to root 2 by 3 into c. Now we have taken z is equal to 2x plus 3y plus 1. Now here putting z is equal to 2x plus 3y plus 1. This implies tan inverse root 3 by 2 into 2x plus 3y plus 1 the whole is equal to root 6 into x plus a. Now this implies root 3 by 2 into 2x plus 3y plus 1 the whole is equal to tan root 6 into x plus a the whole. And this implies 2x plus 3y plus 1 is equal to root 2 by 3 into tan root 6 into x plus a the whole which is the required solution. So in this way we can solve the differential equations in which the variables are not separable. So in this session you have learnt about the equation reducible to variables separable and this completes our session hope you all have enjoyed the session.