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How can I find a solution of the following differential equation: $$\frac{d^2y}{dx^2} =\exp(x^2+ x)$$

Thanks!

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Alpha obtains an answer in term of (imaginary) error function (this means that the primitive of the primitive of $e^{x^2+x}$ is not much more complicated than the primitive itself and that no elementary solution exists). –  Raymond Manzoni Dec 20 '12 at 10:19
    
You can also integrate the Taylor series termwise. –  flavio Dec 20 '12 at 10:33

3 Answers 3

Let $F$ be a primitive of $x \mapsto e^{x^2+x}$ and let $G$ be a primitive of $F$. Then $$y(x)=G(x)+C_1 x + C_2.$$

Remark. My answer is no joke: no elementary expression can be given to $F$ and $G$.

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$$\frac{d^2y}{dx^2}=f(x)$$ Integrating both sides with respect to x, we have $$\frac{dy}{dx}=\int f(x)~dx+A=\phi(x)+A$$ Integrating again $$y=\int \phi(x)~dx+Ax+B=\chi(x)+Ax+B$$

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$\dfrac{d^2y}{dx^2}=e^{x^2+x}$

$\dfrac{dy}{dx}=\int e^{x^2+x}~dx$

$\dfrac{dy}{dx}=\int_k^xe^{x^2+x}~dx+C_1$

$y=\int\int_k^xe^{x^2+x}~dx~dx+\int C_1~dx$

$y=\int_k^x\int_k^xe^{x^2+x}~dx~dx+C_1x+C_2$

By http://mathworld.wolfram.com/RepeatedIntegral.html,

$y=\int_k^x(x-t)e^{t^2+t}~dt+C_1x+C_2$

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