What is the solution for (dy)/(dx)=x^2-1,y(0)=1 ?

The solution for (dy)/(dx)=x^2-1,y(0)=1 is y=(x^3)/3-x+1

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

\frac{d y}{d x} = x^{2} - 1, y (0) = 1

y = \frac{x ^{3}}{3} - x + 1

Solution steps

\frac{d y}{d x} = x^{2} - 1

Substitute

\frac{d y}{d x}

with

y^{'} (x)

y^{^{'}} (x) = x^{2} - 1

f^{^{'}} (x) = g (x)

then

f (x) = \int g (x) d x

y = \int x^{2} - 1 d x

\int x^{2} - 1 d x = \frac{x ^{3}}{3} - x + c_{1}

y = \frac{x ^{3}}{3} - x + c_{1}

Apply initial conditions:

y = \frac{x ^{3}}{3} - x + 1

y = \frac{x ^{3}}{3} - x + 1

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\int_{0}^{x} t^{2} d t

What is the solution for (dy)/(dx)=x^2-1,y(0)=1 ?
The solution for (dy)/(dx)=x^2-1,y(0)=1 is y=(x^3)/3-x+1