Answer:
[tex]y=\sqrt[3]{3x^2+1}[/tex]
Step-by-step explanation:
Given differential equation,
[tex]xdx-y^2dy=0[/tex]
[tex]\implies y^2dy=xdx[/tex]
Integrating both sides,
[tex]\int y^2 dy=\int xdx[/tex]
[tex]\frac{y^3}{3}=x^2+C[/tex]
We have, y(0) = 1,
That is, y = 1 when x = 0,
[tex]\implies \frac{1}{3}=0+C\implies C = \frac{1}{3}[/tex]
[tex]\implies \frac{y^3}{3}=x^2+\frac{1}{3}[/tex]
[tex]y^3=3x^2+1[/tex]
[tex]\implies y=\sqrt[3]{3x^2+1}[/tex]
