Question

X dx − y^2 dy = 0, y(0) = 1

Answer 1

`(y^3)/(3)\ =\ (x^2)/(2)+(1)/(3)`

Explanation:

According to the given question,

The given differential equation is

`xdx\ -\ y^2dy\ =\ 0,\ y(0)\ =\ 1`

`=>\ xdx\ =\ y^2dy`

On integrating both sides, we will have

`=>\ \int{xdx}\ =\ \int{y^2dy}`

`=>\ (x^2)/(2)\ =\ (y^3)/(3)+C` (1)

Now, according to question

y(0)=1

so, we can write

`=>\ (x^2)/(2)\ =\ (y^3)/(3)+C`

`=>\ (0^2)/(2)\ =\ (1^3)/(3)+C`

`=>\ C\ =\ -(1)/(3)`

Now, by putting the value of C in equation (1), we will get

`(x^2)/(2)\ =\ (y^3)/(3)-\ (1)/(3)`

`=>(y^3)/(3)\ =\ (x^2)/(2)+(1)/(3)`

So, the solution of given differential equation will be

`(y^3)/(3)\ =\ (x^2)/(2)+(1)/(3)`