Question

Solve the following differential equation, hence find the particular solution when x = 0, y = 1

`y^3-dy/dx=xdy/dx`

Solution:

`y^3=xdy/dx+dy/dx`

∴ y³ = (x + 1)·`square`

∴ (x + 1)dy = y³ dx

Separating the variables, we get

`1/y^3dy=1/(x+1)dx`

Now integrating, we get

∴ `1/y^3dy=1/(x+1)dx`

∴ `-1/(2y^2)=square+c`   ...(i)

which is required general solution

put x = 0 and y = 1 in (i)

`-1/(2(1)^2)=log|0 + 1| + c`

∴ `square=c`

∴ `-1/(2y^2)=square-1/2`

is the particular solution.

Answer 1

`y^3=xdy/dx+dy/dx`

∴ y³ = (x + 1)·\[\frac{\boxed{dy}}{\boxed{dx}}\] 

∴ (x + 1)dy = y³ dx

Separating the variables, we get

`1/y^3dy=1/(x+1)dx`

Now integrating, we get

∴ `1/y^3dy=1/(x+1)dx`

∴ `-1/(2y^2)` = \[\boxed{log|x+1|}\] + c   ...(i)

which is required general solution

put x = 0 and y = 1 in (i)

`-1/(2(1)^2)=log|0 + 1| + c`

∴ \[\frac{\boxed{{-}1}}{\boxed{2}}\] = c

∴ `-1/(2y^2)` = \[\boxed{log|x+1|}\]`-1/2`

is the particular solution.