Question

x dx − y^2 dy = 0, y(0) = 1

Answer 1

Answer:

$\dfrac{y^3}{3}\ =\ \dfrac{x^2}{2}+\dfrac{1}{3}$

Step-by-step 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

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

$=>\ \dfrac{x^2}{2}\ =\ \dfrac{y^3}{3}+C$          (1)

Now, according to question

y(0)=1

so, we can write

$=>\ \dfrac{x^2}{2}\ =\ \dfrac{y^3}{3}+C$

$=>\ \dfrac{0^2}{2}\ =\ \dfrac{1^3}{3}+C$

$=>\ C\ =\ -\dfrac{1}{3}$

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

   $\dfrac{x^2}{2}\ =\ \dfrac{y^3}{3}-\ \dfrac{1}{3}$

$=>\dfrac{y^3}{3}\ =\ \dfrac{x^2}{2}+\dfrac{1}{3}$

So, the solution of given differential equation will be

$\dfrac{y^3}{3}\ =\ \dfrac{x^2}{2}+\dfrac{1}{3}$