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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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    You can also integrate the Taylor series termwise.2012-12-20

3 Answers 3

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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,

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