Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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


share|cite|improve this question

closed as off-topic by Thomas, William, Davide Giraudo, G. Sassatelli, Jonas Meyer Jul 10 at 20:37

This question appears to be off-topic. The users who voted to close gave this specific reason:

  • "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Thomas, William, Davide Giraudo, G. Sassatelli, Jonas Meyer
If this question can be reworded to fit the rules in the help center, please edit the question.

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

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 $G$.

share|cite|improve this answer

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

share|cite|improve this answer


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


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




share|cite|improve this answer

Not the answer you're looking for? Browse other questions tagged or ask your own question.