 Hello and welcome to the session. I am Deepika here. Let's discuss a question which says Which of the following is a homogeneous differential equation? a 4x plus xy plus 5 into dy minus 3y plus 2x plus 4 into dx is equal to 0 v xy dx minus xq plus yq into dy is equal to 0 c xq plus 2 y square into dx plus 2 xy gy is equal to 0 d y square dx plus x square minus xy minus y square into dy is equal to 0 Now we know that a homogeneous differential equation is of the form dy by dx is equal to g of y over x or dx by dy is equal to h of x over y where g of y over x and h of x over y are homogeneous function of degree 0 So this means that if a differential equation can be written in the form of a homogeneous function of degree 0 then her differential equation is a homogeneous differential equation So in this question we will find out which of the given equations can be written in this form So this is the key idea behind that question We will take the help of this key idea to solve the above question So let's start the solution Now in part a the given differential equation is 4x plus 6y plus 5 into dy minus 3y plus 2x plus 4 into dx is equal to 0 or dy by dx is equal to 3y plus 2x plus 4 over 4x plus 6y plus 5 Now we will try to make this differential equation in the form of a function of y over x So let us divide the numerator and the denominator of right hand side by x So on dividing the numerator and the denominator right hand side of the above equation by x we get dy by dx is equal to 3y over x plus 2 plus 4 over x upon 4 plus 6 into y over x plus 5 over x right hand side of this equation is not of the form g of y over x and so it is not a homogeneous function of degree 0 thus the given equation that is the equation of part a is not a homogeneous differential equation Let us move to the part b in part b the given differential equation is xydx minus xq plus yq into dy is equal to 0 or dy by dx is equal to xq plus yq over xy or we can say dy by dx is equal to x square over y plus y square over x Now since right hand side of above equation is not of the form g of y over x and so it is not a homogeneous function of degree 0 therefore the given differential equation which is given in part b is not a homogeneous differential equation Now let us move to the part c in part c the given differential equation is xq plus 2y square into dx plus 2xy dy is equal to 0 dy by dx is equal to minus of xq plus 2y square over 2xy Let us give this equation as number one Now if we express the right hand side of this differential equation as a homogeneous function of degree 0 then this differential equation is a homogeneous differential equation for this let f of xy is equal to minus xq plus 2y square over 2xy Now if we replace x and y by lambda x and lambda y for any non zero constant lambda we get f of lambda x lambda y is equal to minus lambda q xq plus 2 lambda square y square over 2 lambda square xy Again we can rewrite this function as f of lambda x lambda y is equal to minus let us take lambda square common we have lambda xq plus 2y square upon lambda square into 2xy but this is not equal to lambda raised to power 0 into f of xy so f of xy cannot be expressed in the form of a homogeneous function of degree 0 so we can say the right hand side of equation 1 is not a homogeneous function of degree 0 hence the given equation that is the equation given in part c is not a homogeneous differential equation let us move to the part d in part d the given differential equation is y square dx plus x square minus xy minus y square into dy is equal to 0 now we can rewrite this differential equation as or dy by dx is equal to minus y square over x square minus xy minus y square let us take this equation as number 1 so again if we express the right hand side of this differential equation as a homogeneous function of degree 0 then this differential equation is a homogeneous differential equation for this let f of xy is equal to minus y square over x square minus xy minus y square now if we replace x and y by lambda x and lambda y for any non-zero constant lambda we have f of lambda x lambda y is equal to minus lambda square y square over lambda square x square minus lambda square xy minus lambda square y square again we can rewrite this function as let us f of lambda x lambda y is equal to lambda square into minus y square over lambda square into x square minus xy minus y square now this can be written as lambda raised to power 0 into f of xy thus the right hand side of equation 1 is a homogeneous function of degree 0 therefore the given differential equation that is the equation given in power t is a homogeneous differential equation hence we have seen that only the equation given in power t is a homogeneous differential equation so the answer for the above question is d I hope the solution is clear to you and you have enjoyed the session bye and take care