Initial Value problem (differential calculus)?

Solve the initial value problem

dy/dx = x^2 - 1, y(0) = 1

This is a separate variables where x and y can be integrated separately.

So

dy/dx = x^2 - 1

Becomes

dy = (x^2-1) dx

Now integrate both sides

Y = x^3/3 - x +C , where C is the constant of integration.

We know that y(0)=1

So putting these values in we get

1 = C

So y= x^3/3 - x +1

y' = x^2 - 1

Indefinite integration gives infinite solutions

y = (1/3)x^3 - x + C (°)

to get the unique solution we need one more information

y(0) = 1

means that for x = 0, y = 1

that is: substitute x = 0 and y = 1 in (°) and solve for C

1 = (1/3)· 0^3 - 0 + C

C = 1

the value of C we've found must be substituted in the general equation (°)

y = (1/3)x³ - x + 1

the actual values for y are

at x = 1, y = 1/3

at x = 2, y = 5/3

Calculus Initial Value Problem

dy/dx = x^2 - 1,

=>dy=(x^2-1)dx

=>y=x^3/3-x+c

=>1=0+0+c

=>y=x^3/3-x+1

@ x=1

=>y=1/3

@x=2

=>y=8/3-2+1

=5/3

y=x^3/3-x+C, C=1

y=X^3/3-X+1

1/3-1+1=1/3

8/3-6/3+3/3=5/3=1&2/3