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}

y = 1 + (2 x + 1)^{8}

\int (x^{2} + 1) n x d x

\int_{8}^{R} \frac{1}{x ^{2} - 9 x + 18} d x

f (x) = \frac{6}{3 x + 1}, (- 1, - 3)

\int \frac{1}{2} ln (\frac{1 + x}{1 - x}) d 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}