Se #(dy)/(dx) = x^2 - 2x - 4# com a condição inicial #y(3) = -6#, resolva para #y#?

#y = x^3/3 - x^2 - 4x - 6#

I don't really know Portuguese but here goes.

Given:

#(dy)/(dx) = x^2 - 2x - 4#

and initial condition:

#y(3) = -6#

What we first do is separation of variables (separação de variáveis). That is, we get our #x# and #y# variables isolated on each side.

#dy = x^2 - 2x - 4dx#

Then:

#int dy = int x^2 - 2x - 4 dx#

The arbitrary integration constant (constante de integração) #C# can be placed on either side.

#y = x^3/3 - x^2 - 4x + C#

Now we can apply the initial condition (inicial indicada) to get what the constant #C# is.

#y(3) = 6#

#= (6)^3/3 - (6)^2 - 4(6) + C#

Thus:

#6 = 216/3 - 36 - 24 + C#

#=> C = -6#

Therefore:

#color(blue)(y = x^3/3 - x^2 - 4x - 6)#

CHECK

Is #y# a solution to the ordinary differential equation (equação diferencial ordinária)? (we assume uniqueness.)

#(dy)/(dx) stackrel(?" ")(=) x^2 - 2x - 4#

#= d/(dx)[x^3/3 - x^2 - 4x - 6] = x^2 - 2x - 4# #color(blue)(sqrt"")#

Is the initial condition (inicial indicada) satisfied?

#y(3) = (6)^3/3 - (6)^2 - 4(6) - 6 stackrel(?" ")(=) 6#

#= 216/3 - 36 - 24 - 6#

#= 72 - 36 - 24 - 6#

#= 6# #color(blue)(sqrt"")#