 Hello and welcome to the session. In this session we will discuss about the methods of solving first order first degree differential equations. First we have differential equations with variables separable. Variable separable method is used to solve such an equation in which variables can be separated completely. That is terms containing y should remain with dy and terms containing x should remain with dx. A first order first degree differential equation is of the form dy upon dx is equal to f of xy. Now if f of xy can be expressed as a product of gx and hy where gx is a function of x and hy is a function of y then this differential equation is said to be a variable separable type. And so this can be written as dy upon dx is equal to gx into hy. Now if hy is not equal to 0 then separating the variables this can be written as 1 upon hy dy is equal to gx dx. Now integrating both the sides we get integral 1 upon hy dy is equal to integral gx dx. This provides the solution of the given differential equation in the form hy is equal to gx plus c where hy is the anti derivative of 1 upon hy and gx is anti derivative of gx and the c is the arbitrary constant. Let's try and solve the differential equation dy upon dx plus y equal to 1 by variable separable method. This can be rewritten as dy upon dx equal to 1 minus y or dy upon dx is equal to minus of y minus 1. Now separating the variables we get dy upon y minus 1 is equal to minus dx. Integrating both the sides that is integral dy upon y minus 1 is equal to integral minus dx. This gives us log modulus y minus 1 is equal to minus x plus c or log modulus y minus 1 plus x equal to c where the c is the arbitrary constant. So this is the required solution of the given differential equation. Next we shall discuss about homogeneous differential equations. A differential equation which can be expressed in the form dy upon dx equal to f of xy or dx upon dy equal to g of xy where f of xy and g of xy are homogeneous functions of degree 0. So equations of these kinds are homogeneous differential equations. If we have a homogeneous differential equation of the type dy upon dx equal to f of xy equal to g of y upon x then first of all we substitute y upon x equal to v or we can say y is equal to vx. Then on differentiating with respect to x we get dy by dx equal to v plus x dv by dx. Substituting this value of dy upon dx in this differential equation we get v plus x dv by dx equal to g of v that is we get x dv by dx equal to gv minus v. Now on separating the variables in this equation we get 1 upon gv minus v dv is equal to dx upon x. Now integrating both sides of this equation we get integral dv upon gv minus v equal to integral dx upon x plus c. This gives the general solution of the given differential equation by substituting v equal to y upon x. Let's try and solve the differential equation x dy by dx equal to x plus y. First we divide both the sides by x we get dy by dx equal to 1 plus y upon x. Now we substitute y upon x as v that is y is equal to vx. Now differentiating both the sides with respect to x we get dy upon dx is equal to v plus x dv by dx. Now we substitute this value of dy upon dx in this differential equation this gives us v plus x dv by dx is equal to 1 plus y upon x which is equal to v so this is equal to 1 plus v which gives us x dv by dx is equal to 1. Now on separating the variables we get dv is equal to dx upon x. Now we integrate both the sides so we get integral dv is equal to integral dx upon x that is we have v is equal to log modulus x plus c. We substitute the value for v as y upon x so this gives us y upon x is equal to log modulus x plus c or this could also be written as y upon x minus log modulus x equal to c. This is the required solution of the given differential equation where the c is the arbitrary constant. Next we have linear differential equation a differential equation of the form dy upon dx plus py equal to q where p and q are constants or functions of x only then the differential equation of this kind is called the first order linear differential equation. If a differential equation is of this kind then the first step is to find the integrating factor that is if which is given by e to the power integral p dx then the solution of this differential equation is given by y multiplied by the integrating factor if equal to integral q multiplied by the integrating factor if dx plus c. If the differential equation is of the form dx upon dy plus p1x equal to q1 where p1 and q1 are constants which is the functions of y only then in this case the integrating factor if would be given by e to the power integral p1 dy and the solution would be given by x into the integrating factor if equal to integral q1 multiplied by the integrating factor if dy plus c. Let's try and find the solution for the differential equation dy upon dx plus y cot x is equal to 2 cos x. This is of the form dy upon dx plus py equal to q here we have p is equal to cot x and q is equal to 2 cos x so the integrating factor if in this case is given by e to the power integral p dx that is cot x dx which is equal to e to the power log sin x and so integrating factor is equal to sin x. Now the required solution is given by y multiplied by the integrating factor sin x equal to integral q that is 2 cos x multiplied by the integrating factor sin x dx plus c1 we have y sin x is equal to integral sin 2x dx plus c1 which is equal to minus cos 2x upon 2 plus c1 hence we get 2y sin x is equal to minus cos 2x plus 2c1 or this could be written as 2y sin x plus cos 2x is equal to c where c is equal to 2c1 is an arbitrary constant so this is the required solution of the given differential equation. So this completes the session hope you understood all the three methods of solving a differential equation.