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Question
Solve the differential equation `ye^(x/y) dx = (xe^(x/y) + y^2)dy, (y != 0)`
Solution 1
`ye^(x/y)dx = (xe^(x/y) + y^2)dy`
`\implies ye^(x/y) dx/dy = xe^(x/y) + y^2`
`\implies e^(x/y) [y.dx/dy - x]` = y2
`\implies e^(x/y). ([y.dx/dy - x])/y^2` = 1 ...(1)
Let `e^(x/y)` = z.
Differentiating it with respect to y, we get:
`d/dy(e^(x/y)) = dz/dy`
`\implies e^(x/y) . d/dy (x/y) = dz/dy`
`\implies e^(x/y). [(y.dx/dy - x)/y^2] = dz/dy` ...(2)
From equation (1) and equation (2), we get:
`dz/dy` = 1
`\implies` dz = dy
Integrating both sides, we get:
z = y + C
`\implies e^(x/y)` = y + C
Solution 2
`ye^(x/y)dx = (xe^(x/y) + y^2)dy`
`\implies e^(x/y) (ydx - xdy)` = y2 dy
`\implies e^(x/y) ((ydx - xdy)/y^2)` = dy
`\implies e^(x/y)d(x/y)` = dy
`\implies int e^(x/y)d (x/y) = int dy`
`\implies e^(x/y)` = y + c, where 'c' is an arbitrary constant of integration.
Solution 3
We have, `dx/dy = (xe^(x/y) + y^2)/(y.e^(x/y)`
`\implies dx/dy = x/y + y/e^(x/y)` ...(i)
Let x = vy `\implies dx/dy = v + y.(dv)/dy`;
So equation (i) becomes `v + y (dv)/dy = v + y/e^v`
`\implies y (dv)/dy = y/e^v`
`\implies` ev dv = dy
On integrating we get,
`inte^v dv = int dy`
`\implies` ev = y + c
`\implies` ex/y = y + c
where 'c' is an arbitrary constant of integration.
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