Question

Find the particular solution of the differential equation:

2y e^x/y dx + (y - 2x e^x/y) dy = 0 given that x = 0 when y = 1.

Answer 1

`2ye^(x/y)dx+(y-2xe^(x/y))dy=0`

`=>dx/dy=(2xe^(x/y-y))/(2ye^(x/y))`

Given differential equation is a homogeneous differential equation.

∴ Put x = vy

`dx/dy=v+y (dv)/dy`

`v+y(dv)/dy=(2ve^v-1)/(2e^v)`

`=>y(dv)/dy=(2ve^v-1)/(2e^v)-v`

`=>y(dv)/dy=-1/(2e^v)`

`=>2e^vdv=-1/ydy`

Integrating on both the sides

`=>2inte^vdv=-int1/ydy`

`=>2e^v=-log|y|+logC`

`=>2e^v=log|c/y|`

`=>2e^(x/y)=log|c/y|`

Given that at x = 0, y = 1

`2e^0= log|c/1|`

⇒ C = e²

`:.2e^(x/y)=log""e^2/y`

`=>logy=-2e^(x/y)+2`

`=>y=e^2-2e^(x/y)`