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

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

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y^'=x^2-1,y(0)=1

y^{^{'}} = x^{2} - 1, y (0) = 1

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

Solution steps

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

\int \frac{1}{289 + x ^{2}} d x

\int_{0}^{1} (x^{15} + 1 5^{x}) d x

x^{2} y^{^{''}} - x y^{^{'}} + y = x

x \to 9 - lim (x^{2} - 3 x)

\int \frac{( ln ( x ) ) ^{\frac{7}{2}}}{x} d x

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