Given #x^t * y^m - (x+y)^(m+t)=0# determine #dy/dx# ? Calculus Applications of Definite Integrals Solving Separable Differential Equations 1 Answer Cesareo R. Oct 28, 2016 See below. Explanation: #f(x,y)=x^t * y^m - (x+y)^(m+t)=0# #df = f_xdx+f_ydy=0# so #dy/dx=-f_x/(f_y)# but #f_x=t/x x^ty^m-(t+m)/(x+y)(x+y)^(t+m)# and #f_y=m/y x^ty^m-(t+m)/(x+y)(x+y)^(t+m)# but #x^ty^m=(x+y)^(t+m)# then #f_x=(t/x-(t+m)/(x+y))x^ty^m# and #f_y=(m/y-(t+m)/(x+y))x^ty^m# so #dy/dx=-(t/x - (t + m)/(x + y))/(m/y - (t + m)/(x + y))=y/x# Answer link Related questions How do you solve separable differential equations? How do you solve separable first-order differential equations? How do you solve separable differential equations with initial conditions? What are separable differential equations? How do you solve the differential equation #dy/dx=6y^2x#, where #y(1)=1/25# ? How do you solve the differential equation #y'=e^(-y)(2x-4)#, where #y5)=0# ? How do you solve the differential equation #(dy)/dx=e^(y-x)sec(y)(1+x^2)#, where #y(0)=0# ? How do I solve the equation #dy/dt = 2y - 10#? Given the general solution to #t^2y'' - 4ty' + 4y = 0# is #y= c_1t + c_2t^4#, how do I solve the... How do I solve the differential equation #xy'-y=3xy, y_1=0#? See all questions in Solving Separable Differential Equations Impact of this question 1399 views around the world You can reuse this answer Creative Commons License