Question

x dy/dx + ylogy = xye^(x)​

Answer 1

Given, x(dy/dx)+y log y=xy e^x

divide whole eqn by y:

x/ydy/dx+log y=x.e^x

let xlog y=t;

x/y dy/dx+logy=dt/dx

replacing in original equation

weget: dt/dx=x.e^x

then dt=x.e^xdx

integrating bothsides:

∫dt=∫x.e^xdx

t=x.e^x−∫e^xdx (Applying by parts intergation )

t =x e^x-e^x+C

t=e^x(x−1)+C

placing value of t, we get,

xlogy=e^x (x−1)+C

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