# Stuck trying to integrate e^-(x^2)/2

## Homework Statement

find general solution of

dy/dx=xy+x

using integrating factor

## The Attempt at a Solution

rearange

dy/dx-xy=x

take integrating factor as
e^int(-x)=e^-(x^2)/2

multiply throughout

dy/dx(e^-(x^2)/2)y=(e^-(x^2)/2)x

integrate

e^-(x^2)/2*y=int (e^-(x^2)/2)*x dx

dont know how to integrate (e^-(x^2)/2)*x

cheers

dooogle

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Mark44
Mentor

## Homework Statement

find general solution of

dy/dx=xy+x

using integrating factor

## The Attempt at a Solution

rearange

dy/dx-xy=x

take integrating factor as
e^int(-x)=e^-(x^2)/2

multiply throughout

dy/dx(e^-(x^2)/2)y=(e^-(x^2)/2)x

integrate

e^-(x^2)/2*y=int (e^-(x^2)/2)*x dx

dont know how to integrate (e^-(x^2)/2)*x
An ordinary substitution can be used in this integral. Luckily you have that factor of x in the integrand.
$$\int xe^{\frac{-x^2}{2}} dx$$

cheers for the help

i let u=-(x^2)/2

du/dx=-2x/2=-x

du=-x dx

so -int e^u du = int e^-(x^2)/2 dx

=-e^u =-e^-(x^2)/2 +c

dooogle

Mark44
Mentor
A few corrections...
i let u=-(x^2)/2

du/dx=-2x/2=-x

du=-x dx

so -int e^u du = int e^-(x^2)/2 dx
It's the other way around, plus you left off the factor of x in the integral on the right.
$$\int xe^{\frac{-x^2}{2} dx ~=~ -\int e^u~du$$
=-e^u =-e^-(x^2)/2 +c
In the line above, the arbitrary constant comes in as soon as you have the antiderivative. Also, your 2nd exponential expression is not quite right.

This line should be
=-e^u + C = e^(-(x^2)/2) + C