Initial Value Differential Equation

[goku900]

I have dy/dx = x^2/y^2

=y^2 dy = x^2 dx

∫y^2 dy = ∫x^2 dx

y^3/3 = x^3/3+ C

y^3 = x^3 + 3C

3C= some constant k

y = (x^3+k)^(1/3)

I now have to solve the initial value problem of y(0)=2 ; kinda forgot how to do this part.

 

[HallsofIvy]

I have dy/dx = x^2/y^2

=y^2 dy = x^2 dx

∫y^2 dy = ∫x^2 dx

y^3/3 = x^3/3+ C

y^3 = x^3 + 3C

3C= some constant k

y = (x^3+k)^(1/3)

I now have to solve the initial value problem of y(0)=2 ; kinda forgot how to do this part.

Seriously? "y(0)= 2" means "when x= 0, y= 2" so replace x and y with 0 and 2, respectively, and solve for k.

 

[DrPhil]

I have dy/dx = x^2/y^2

y^2 dy = x^2 dx.........omitted "=" you had at beginning of line

∫y^2 dy = ∫x^2 dx

y^3/3 = x^3/3+ C

y^3 = x^3 + 3C

3C= some constant k

y = (x^3+k)^(1/3)

I now have to solve the initial value problem of y(0)=2 ; kinda forgot how to do this part.

Substitute x=0 and y(0)=2 in this equation, and solve for k. That ism the initial value lets you find the constant of integration.