What is the solution for y^{''}=x^2-1 ?

The solution for y^{''}=x^2-1 is y=(x^4}{12}-\frac{x^2)/2+c_{1}x+c_{2}

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

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

y = \frac{x ^{4}}{12} - \frac{x ^{2}}{2} + c_{1} x + c_{2}

Solution steps

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

Solve

y^{''} (x) = x^{2} - 1 : y^{'} (x) = \frac{x ^{3}}{3} - x + c_{1}

Solve

y^{'} (x) = \frac{x ^{3}}{3} - x + c_{1} : y = \frac{x ^{4}}{12} - \frac{x ^{2}}{2} + c_{1} x + c_{2}

y = \frac{x ^{4}}{12} - \frac{x ^{2}}{2} + c_{1} x + c_{2}

5^{x}

\int 3 x - 4 d x

x \to - 1 lim (\frac{x - 2}{x ^{2} - 1})

\int \frac{1}{u ^{2} + 1} d u

\frac{d y}{d x} = \frac{1}{x}

What is the solution for y^{''}=x^2-1 ?
The solution for y^{''}=x^2-1 is y=(x^4}{12}-\frac{x^2)/2+c_{1}x+c_{2}