Question

solve the general equation y'=x^2y​

Answer 1

Step-by-step explanation:

dy/dx = x^2 * y

1/y dy = x^2 dx

integrating both sides, we get

log y = x^3 /3

Answer 2

$\mathfrak{\huge{\pink{\underline{\underline{AnSwEr:-}}}}}$

Actually Welcome to the Concept of the Diffrential Equations

So after solving we get as,

Since here ,

y' = dy/dx

so we get as

dy/dx = x^2y

dy/y = x^2 dx

so integrating both sides we get as

log y = x^3/3 + C

where C is the integral constant