dy/(y^2y) = dx but 1/(y^2y) = 1/(y1) 1/y, so (1/(y1) 1/y)dy = dx ln(y1) ln y = xk ln ((y1)/y) = xk (y1)/y = e^(xk) = e^k * e^x = ce^x y1 = cye^x y(1ce^x) = 1 y = 1/(1ce^x) Hmmm Wolframalpha says y = ce^x/(e^cx1) but they are the same, since u/(u1) = 1 1/(u1) so there's just a different constant c Still, better doublecheck my math Transcript Ex 96, 10 For each of the differential equation given in Exercises 1 to 12, find the general solution ( ) / =1 Step 1 Put in form / Py = Q or / P1x = Q1 (x y) / = 1 Dividing by (x y), / = 1/(( )) / = ( ) / x = / ( 1) x = Step 2 Find P1 and Q1 Comparing (1) with / P1x = Q1 P1 = 1 and Q1 = y Step 3 Find Integrating factor, IFView Ejercicios Unidad 1docx from ITI 1234 at Polytechnic University of Victoria, Ciudad Victoria EJEMPLOS METODO DE VARIABLE SEPARABLES y '=xy dy =xy dx dy =xdx y ∫ dy = xdx y ∫ ln
How To Solve X Y 1 Dy X Y 1 Dx Quora
(y ln x)^-1 dy/dx=(x/y+1)^2