Solve differential equations using the given conditions?

y'=(1/y^2) ; y(1)=1

y^2dy = dx

∫ y^2dy = ∫ dx

1/3y^3 = x + C

when y(1) = 1

1/3(1) = 1 + C

1/3 - 1 = C

C = -2/3

[1/3y^3 = x - 2/3]3

y^3 = 3x - 2

y = ∛(3x - 2) answer//

There is no need for brackets around the right hand side of the equation.

Find the general solution by separating the variables then integrating:

y' = 1 / y²

dy / dx = 1 / y²

y² dy = dx

∫ y² dy = ∫ 1 dx

y³ / 3 = x + C

y³ = 3x + C

y = ∛(3x + C)

Find the particular solution by solving for the constant:

When x = 1, y = 1

∛(3 + C) = 1

3 + C = 1

C = -2

y = ∛(3x - 2)

you can make 1/y^2 to just y^-2.

integrate y^-2 to get -y^-1. rewrite -1/y + C. then solve for c.

-1/1 + C = 1

-1+C=1

C=2

particular solution: -1/y + 2.