# Thread: integration of e^x with function in x

1. ## integration of e^x with function in x

Sir,

while solving a differential equation problem

I got the equation as y/(x+1) = integration of e3x(x + 1) + c (integration constant)

what I did in the integration I multiplied and divided by 3

3e3x(x + 1)
_________

3

e3x(3x + 3) I took f(x) = 3x and f'(x) = 3]
_________

3

I used integration of ex [f(x) + f'(x)]

= ex f(x) + c

= xex + c
---
3

But they have solved the problem by using parts

e3x(x + 2)
_________ + c

9

Kindly explain were I went wrong?

with regards,

Aranga

2. ## Re: integration of e^x with function in x

Originally Posted by arangu1508
Sir,

while solving a differential equation problem

I got the equation as y/(x+1) = integration of e3x(x + 1) + c (integration constant)
.
.
.
I would like you to post the complete original problem, before you started working on it, because it sounds like you gave
us your version of part of the problem along the way without the original problem statement.

3. ## Re: integration of e^x with function in x

solve the differential equation

(x + 1)y' - y = [e^(3x)] [(x + 1)^3]

4. ## Re: integration of e^x with function in x

Originally Posted by arangu1508
Sir,

while solving a differential equation problem

I got the equation as y/(x+1) = integration of e3x(x + 1) + c (integration constant)

what I did in the integration I multiplied and divided by 3

3e3x(x + 1)
_________

3

e3x(3x + 3) I took f(x) = 3x and f'(x) = 3]
_________

3

I used integration of ex [f(x) + f'(x)]

wouldn't this be

$\color{red} {e^{f(x)} [f(x) + f'(x)]}$ ?
you would still need to integrate by parts, no?

5. ## Re: integration of e^x with function in x

Originally Posted by arangu1508
solve the differential equation

(x + 1)y' - y = [e^(3x)] [(x + 1)^3]
I'd begin by writing:

$\displaystyle y'-\frac{1}{x+1}y=e^{3x}(x+1)^2$

Compute the integrating factor:

$\displaystyle \mu(x)=\exp\left(-\int\frac{1}{x+1}\,dx\right)=\frac{1}{x+1}$

And the ODE becomes:

$\displaystyle \frac{1}{x+1}y'-\frac{1}{(x+1)^2}y=e^{3x}(x+1)$

Or:

$\displaystyle \frac{d}{dx}\left(\frac{y}{x+1}\right)=e^{3x}(x+1)$

This agrees with what you did, and as stated integration by parts can be used to integrate the RHS:

$\displaystyle u=x+1\implies dx=dx$

$\displaystyle dv=e^{3x}\,dx\implies v=\frac{1}{3}e^{3x}$

Hence:

$\displaystyle \frac{y}{x+1}=\frac{x+1}{3}e^{3x}-\frac{1}{3}\int e^{3x}\,dx$

$\displaystyle \frac{y}{x+1}=\frac{x+1}{3}e^{3x}-\frac{1}{9}e^{3x}+C$

And so:

$\displaystyle y(x)=\frac{x+1}{9}\left(3(x+1)e^{3x}-e^{3x}+c_2\right)$

$\displaystyle y(x)=\frac{x+1}{9}\left((3x+2)e^{3x}+c_1\right)$